In the previous section, we saw the case of a car in which the real wheels are self propagating. We also saw the case of a cart whose wheels need external force to rotate
1. Consider an independent circular disc. The word 'independent' is specified to clarify that, the disc has no external connections like to an axle or an engine
• It is kept at a height away from the ground. This is shown in fig.7.145 (a) below:
2. Assume that this disc has the ability to self propagate. Like a rear car wheel
• So let it spin. When it has attained considerable angular speed, say $\mathbf\small{\vec{\omega}_i}$, stop the engine
3. If there is no frictional resistance or air resistance, that disc will continue to spin for ever with angular velocity $\mathbf\small{\vec{\omega}_i}$
• We know the reason:
If there is no external torque, angular momentum $\mathbf\small{I\vec{\omega}_i}$ will be conserved
4. Lower the disc slowly so that the disc gently touches the floor
• Care should be taken to ensure that, while placing the disc, no horizontal force is produced on the disc
5. Assume that the floor is a rough surface
• So there will be friction between the disc and the floor
• In the previous section, we have seen that in the case of a self propagating disc, the frictional force will be in the direction of motion
6. We have to determine whether this friction is static or kinetic
(i) The disc is continuously spinning. There is no instant at which any point on the periphery is at rest
(ii) Let the point of contact be P
(iii) Consider the instant at which the contact with the floor takes place. Even at that instant, the particle P is in motion
(iv) If it is to be at rest, the following two velocities must be equal:
♦ The tangential velocity of P, which is towards the left
♦ The velocity of the center of mass C, which is towards the right
(v) But at the instant of touch down, the 'velocity of the center of mass towards the right' is zero
(vi) So the point of contact is not at rest at the instant of touch down
(vii) So the frictional force is that of kinetic friction
(viii) This is shown as $\mathbf\small{|\vec{f_k}|}$ in the fig.b
7. The $\mathbf\small{|\vec{f_k}|}$ is acting on the disc. That means an external force is acting on the disc
• So the center of mass of the disc will have linear acceleration $\mathbf\small{|\vec{a}|}$
• We have: Force = mass × acceleration
So this acceleration can be obtained using the expression: $\mathbf\small{|\vec{a}|=\frac{|\vec{f_k}|}{m}}$
8. Note: For translational motion, we assume that the external net force acts at the center of mass
• So in our present case, the $\mathbf\small{\vec{f_k}}$ can be assumed to be acting at the center of the disc
• So the expression written in (7) is valid
9. Thus we obtained the linear acceleration $\mathbf\small{|\vec{a}|}$
• The initial linear velocity $\mathbf\small{|\vec{v}_i|}$ is zero
• So the linear velocity $\mathbf\small{|\vec{v}_{(t)}|}$ after a time interval of 't' s after touch down will be given by:
$\mathbf\small{|\vec{v}_{(t)}|=|\vec{v}_i|+|\vec{a}|\,t}$
$\mathbf\small{\Rightarrow|\vec{v}_{(t)}|=0+|\vec{a}|\,t}$
$\mathbf\small{\Rightarrow|\vec{v}_{(t)}|=|\vec{a}|\,t}$
10. Next we consider the angular motion
• The force $\mathbf\small{\vec{f_k}}$ acts on the disc
• But $\mathbf\small{\vec{f_k}}$ it does not pass through the center of the disc
• So the disc will experience a torque
• Obviously, this torque is given by: $\mathbf\small{|\vec{\tau}|=|\vec{f_k}|\,R}$
11. If there is a torque, there will be an acceleration $\mathbf\small{|\vec{\alpha}|}$
• This $\mathbf\small{|\vec{\alpha}|}$ can be obtained using the expression: $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}}$
• Recall that $\mathbf\small{\vec{\tau}=I\,\vec{\alpha}}$ corresponds to $\mathbf\small{\vec{F}=m\,\vec{a}}$
12. Thus we obtained the angular acceleration $\mathbf\small{|\vec{\alpha}|}$
• The initial angular speed is $\mathbf\small{\vec{\omega}_i}$
• So the angular speed $\mathbf\small{|\vec{\omega}_{(t)}|}$ after a time interval of 't' s after touch down will be given by:
$\mathbf\small{|\vec{\omega}_{(t)}|=|\vec{\omega}_i|-|\vec{\alpha}|\,t}$
■ Why is a negative sign given to $\mathbf\small{|\vec{\alpha}|}$?
• The answer can be written in 5 steps:
(i) The torque is created by $\mathbf\small{\vec{f_k}}$
(ii) From the direction of $\mathbf\small{\vec{f_k}}$, it is obvious that, the torque opposes the initial angular velocity
(iii) That means, the angular speed will go on decreasing
(iv) That means, there is angular 'deceleration'
(v) So we give a negative sign
13. Let us write the two available equations together:
• From (9), we have: $\mathbf\small{|\vec{v}_{(t)}|=|\vec{a}|\,t}$
• From (12), we have: $\mathbf\small{|\vec{\omega}_{(t)}|=|\vec{\omega}_i|-|\vec{\alpha}|\,t}$
14. The rolling of the disc just after touch down is not 'pure rolling'
♦ We proved this by showing that the velocity of the contact point P is not zero
• We also saw in (12) that, the disc suffers angular deceleration
♦ So a time will come at which the angular velocity becomes so low that, the condition $\mathbf\small{|\vec{v}_{CM}|=R\,|\vec{\omega}|}$ is satisfied
♦ At that time, the rolling changes to pure rolling
Let the pure rolling begin after 't' seconds
• For pure rolling, we have: $\mathbf\small{|\vec{v}_{CM}|=R\,|\vec{\omega}|}$
• All points of the disc will have the same linear velocity $\mathbf\small{|\vec{v}_{CM}|}$
• So we can simply write $\mathbf\small{|\vec{v}|}$ instead of $\mathbf\small{|\vec{v}_{CM}|}$
• So we can write as:
When pure rolling begins: $\mathbf\small{|\vec{v}_{(t)}|=R\,|\vec{\omega}_{(t)}|}$
15. Now we substitute for $\mathbf\small{|\vec{v}_{(t)}|}$ and $\mathbf\small{|\vec{\omega}_{(t)}|}$
• We use the equations in (13) for the substitution. We get:
$\mathbf\small{|\vec{a}|\,t=R\,\left(|\vec{\omega}_i|-|\vec{\alpha}|\,t\right)}$
$\mathbf\small{\Rightarrow|\vec{a}|\,t=R|\vec{\omega}_i|-R|\vec{\alpha}|\,t}$
$\mathbf\small{\Rightarrow(|\vec{a}|+R|\vec{\alpha}|)\,t=R|\vec{\omega}_i|}$
• Thus we get:
Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
• Where:
♦ t is the time after which pure rolling begins
♦ R is the radius of the disc
♦ $\mathbf\small{|\vec{\omega}_i|}$ is the angular velocity at the time of touch down
♦ $\mathbf\small{|\vec{a}|=\frac{|\vec{f_k}|}{m}}$
♦ $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
We will see a solved example:
Solved example 7.46
A wheel spinning at an angular velocity of 70 rad s-1 is slowly lowered on to the ground with out any horizontal force. Radius of the wheel is 0.4 m and it's mass is 15 kg.
(a) After what time will the pure rolling begin?
(b) What is the distance traveled by the wheel before the beginning of pure rolling ?
Coefficient of kinetic friction between the wheel and the ground is 0.5
Take g = 9.8 ms-2
Solution:
Part (a):
1. We have Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
2. Calculation of $\mathbf\small{|\vec{a}|}$:
• $\mathbf\small{|\vec{a}|=\frac{|\vec{F}|}{m}=\frac{|\vec{f_k}|}{m}=\frac{\mu_k\,m\,|\vec{g}|}{m}=\mu_k|\vec{g}|}$
• So we get: $\mathbf\small{|\vec{a}|=\mu_k|\vec{g}|=(0.5 \times 9.8)}$ = 4.9 ms-2
3. Calculation of $\mathbf\small{|\vec{\alpha}|}$:
• $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
• For a disc $\mathbf\small{I=\frac{mR^2}{2}}$
• So we get: $\mathbf\small{|\vec{\alpha}|=\frac{2R\,|\vec{f_k}|}{mR^2}=\frac{2\,|\vec{f_k}|}{mR}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2\,\mu_k\,m\,|\vec{g}|}{mR}=\frac{2\,\mu_k\,|\vec{g}|}{R}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2(0.5)(9.8)}{0.4}=25.4}$ rad s-2.
4. Substituting the results from (2) and (3) in (1), we get:
$\mathbf\small{t= \frac{0.4 \times 70}{[(4.9)+0.4(25.4)]}=1.86\, \rm{s}}$
Part (b):
1. Consider the two instances:
(i) Instance at which the contact with the floor is made
(ii) Instance at which the pure rolling begins
2. From part (a), we have: Time interval between the two instances = 1.86 s
• Linear acceleration with which the wheel travels during this time interval = (0.5 × 9.8) = 4.9 ms-2.
• Initial velocity of this linear motion = 0
• So we can find the velocity at the instance when pure rolling begins
3. We can use the relation $\mathbf\small{|\vec{v}_f|=|\vec{v}_i|+|\vec{a}|\,t}$
• For our present case: $\mathbf\small{|\vec{v}_f|=0+|\vec{a}|\,t}$
$\mathbf\small{\Rightarrow|\vec{v}_f|=|\vec{a}|\,t}$
• Substituting the values, we get: $\mathbf\small{|\vec{v}_f|}$ = (4.9 × 1.86) = 9.114 ms-1.
4. Let $\mathbf\small{|\vec{x}|}$ be the distance traveled by the wheel during this interval
• We can use the relation: $\mathbf\small{|\vec{v}_f|^2-|\vec{v}_i|^2=2|\vec{a}|\,|\vec{x}|}$
• Substituting the values, we get: $\mathbf\small{9.1^2-0=2\times4.9\times|\vec{x}|}$
• Thus we get: $\mathbf\small{|\vec{x}|}$ = 8.45 m
Solved example 7.47
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 𝝅 rad s-1. Which of the two will start to roll earlier ? The co-efficient of kinetic friction is 𝝁k = 0.2
Solution:
• The solid disc and ring are placed simultaneously
• Both have the same initial angular velocity of 10 𝝅 rad s-1
• As soon as they are placed, both will move horizontally in the same direction
• But for both of them, this motion will not be 'pure rolling'
• Pure rolling will start only after some time 't' after the 'instant of touch down'
• This 't' will be different for the two objects
• We have to find which object has the smallest 't'
• Given that the radius is the same 10 cm for both the objects. So we will write:
♦ Radius of disc = Radius of ring = R = 10 cm
1. First we will consider the solid disc
We have:
Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
• Where:
♦ t is the time after which pure rolling begins
♦ R is the radius of the disc
♦ $\mathbf\small{|\vec{\omega}_i|}$ is the angular velocity at the time of touch down
♦ $\mathbf\small{|\vec{a}|=\frac{|\vec{f_k}|}{m}}$
♦ $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
2. Calculation of $\mathbf\small{|\vec{a}|}$:
• $\mathbf\small{|\vec{a}|=\frac{|\vec{F}|}{m}=\frac{|\vec{f_k}|}{m}=\frac{\mu_k\,m\,|\vec{g}|}{m}=\mu_k|\vec{g}|}$
• So we get: $\mathbf\small{|\vec{a}|=\mu_k|\vec{g}|=(0.2 \times 9.8)}$ = 1.96 ms-2
3. Calculation of $\mathbf\small{|\vec{\alpha}|}$:
• $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
• For a disc $\mathbf\small{I=\frac{mR^2}{2}}$
• So we get: $\mathbf\small{|\vec{\alpha}|=\frac{2R\,|\vec{f_k}|}{mR^2}=\frac{2\,|\vec{f_k}|}{mR}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2\,\mu_k\,m\,|\vec{g}|}{mR}=\frac{2\,\mu_k\,|\vec{g}|}{R}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2(0.2)(9.8)}{R}=\frac{3.92}{R}}$ rad s-2.
4. Substituting the results from (2) and (3) in (1), we get:
$\mathbf\small{t_d= \frac{R \times 10 \pi}{[(1.96)+R(\frac{3.92}{R})]}=1.7\pi R\, \rm{s}}$
5. Now we consider the ring
We have:
Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
6. Calculation of $\mathbf\small{|\vec{a}|}$:
• $\mathbf\small{|\vec{a}|=\frac{|\vec{F}|}{m}=\frac{|\vec{f_k}|}{m}=\frac{\mu_k\,m\,|\vec{g}|}{m}=\mu_k|\vec{g}|}$
• So we get: $\mathbf\small{|\vec{a}|=\mu_k|\vec{g}|=(0.2 \times 9.8)}$ = 1.96 ms-2
7. Calculation of $\mathbf\small{|\vec{\alpha}|}$:
• $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
• For a ring $\mathbf\small{I=mR^2}$
• So we get: $\mathbf\small{|\vec{\alpha}|=\frac{R\,|\vec{f_k}|}{mR^2}=\frac{|\vec{f_k}|}{mR}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{\mu_k\,m\,|\vec{g}|}{mR}=\frac{\mu_k\,|\vec{g}|}{R}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{(0.2)(9.8)}{R}=\frac{1.96}{R}}$ rad s-2.
8. Substituting the results from (6) and (7) in (5), we get:
$\mathbf\small{t_r= \frac{R \times 10 \pi}{[(1.96)+R(\frac{1.96}{R})]}=2.55\pi R\, \rm{s}}$
9. Let us write the results together:
• From (4) we have: td = 1.7 𝝅R
• From (8) we have: td = 2.55 𝝅R
• Obviously, td is lesser. That means, the disc will start pure rolling earlier
1. Consider an independent circular disc. The word 'independent' is specified to clarify that, the disc has no external connections like to an axle or an engine
• It is kept at a height away from the ground. This is shown in fig.7.145 (a) below:
 |
| Fig.7.145 |
• So let it spin. When it has attained considerable angular speed, say $\mathbf\small{\vec{\omega}_i}$, stop the engine
3. If there is no frictional resistance or air resistance, that disc will continue to spin for ever with angular velocity $\mathbf\small{\vec{\omega}_i}$
• We know the reason:
If there is no external torque, angular momentum $\mathbf\small{I\vec{\omega}_i}$ will be conserved
4. Lower the disc slowly so that the disc gently touches the floor
• Care should be taken to ensure that, while placing the disc, no horizontal force is produced on the disc
5. Assume that the floor is a rough surface
• So there will be friction between the disc and the floor
• In the previous section, we have seen that in the case of a self propagating disc, the frictional force will be in the direction of motion
6. We have to determine whether this friction is static or kinetic
(i) The disc is continuously spinning. There is no instant at which any point on the periphery is at rest
(ii) Let the point of contact be P
(iii) Consider the instant at which the contact with the floor takes place. Even at that instant, the particle P is in motion
(iv) If it is to be at rest, the following two velocities must be equal:
♦ The tangential velocity of P, which is towards the left
♦ The velocity of the center of mass C, which is towards the right
(v) But at the instant of touch down, the 'velocity of the center of mass towards the right' is zero
(vi) So the point of contact is not at rest at the instant of touch down
(vii) So the frictional force is that of kinetic friction
(viii) This is shown as $\mathbf\small{|\vec{f_k}|}$ in the fig.b
7. The $\mathbf\small{|\vec{f_k}|}$ is acting on the disc. That means an external force is acting on the disc
• So the center of mass of the disc will have linear acceleration $\mathbf\small{|\vec{a}|}$
• We have: Force = mass × acceleration
So this acceleration can be obtained using the expression: $\mathbf\small{|\vec{a}|=\frac{|\vec{f_k}|}{m}}$
8. Note: For translational motion, we assume that the external net force acts at the center of mass
• So in our present case, the $\mathbf\small{\vec{f_k}}$ can be assumed to be acting at the center of the disc
• So the expression written in (7) is valid
9. Thus we obtained the linear acceleration $\mathbf\small{|\vec{a}|}$
• The initial linear velocity $\mathbf\small{|\vec{v}_i|}$ is zero
• So the linear velocity $\mathbf\small{|\vec{v}_{(t)}|}$ after a time interval of 't' s after touch down will be given by:
$\mathbf\small{|\vec{v}_{(t)}|=|\vec{v}_i|+|\vec{a}|\,t}$
$\mathbf\small{\Rightarrow|\vec{v}_{(t)}|=0+|\vec{a}|\,t}$
$\mathbf\small{\Rightarrow|\vec{v}_{(t)}|=|\vec{a}|\,t}$
10. Next we consider the angular motion
• The force $\mathbf\small{\vec{f_k}}$ acts on the disc
• But $\mathbf\small{\vec{f_k}}$ it does not pass through the center of the disc
• So the disc will experience a torque
• Obviously, this torque is given by: $\mathbf\small{|\vec{\tau}|=|\vec{f_k}|\,R}$
11. If there is a torque, there will be an acceleration $\mathbf\small{|\vec{\alpha}|}$
• This $\mathbf\small{|\vec{\alpha}|}$ can be obtained using the expression: $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}}$
• Recall that $\mathbf\small{\vec{\tau}=I\,\vec{\alpha}}$ corresponds to $\mathbf\small{\vec{F}=m\,\vec{a}}$
12. Thus we obtained the angular acceleration $\mathbf\small{|\vec{\alpha}|}$
• The initial angular speed is $\mathbf\small{\vec{\omega}_i}$
• So the angular speed $\mathbf\small{|\vec{\omega}_{(t)}|}$ after a time interval of 't' s after touch down will be given by:
$\mathbf\small{|\vec{\omega}_{(t)}|=|\vec{\omega}_i|-|\vec{\alpha}|\,t}$
■ Why is a negative sign given to $\mathbf\small{|\vec{\alpha}|}$?
• The answer can be written in 5 steps:
(i) The torque is created by $\mathbf\small{\vec{f_k}}$
(ii) From the direction of $\mathbf\small{\vec{f_k}}$, it is obvious that, the torque opposes the initial angular velocity
(iii) That means, the angular speed will go on decreasing
(iv) That means, there is angular 'deceleration'
(v) So we give a negative sign
13. Let us write the two available equations together:
• From (9), we have: $\mathbf\small{|\vec{v}_{(t)}|=|\vec{a}|\,t}$
• From (12), we have: $\mathbf\small{|\vec{\omega}_{(t)}|=|\vec{\omega}_i|-|\vec{\alpha}|\,t}$
14. The rolling of the disc just after touch down is not 'pure rolling'
♦ We proved this by showing that the velocity of the contact point P is not zero
• We also saw in (12) that, the disc suffers angular deceleration
♦ So a time will come at which the angular velocity becomes so low that, the condition $\mathbf\small{|\vec{v}_{CM}|=R\,|\vec{\omega}|}$ is satisfied
♦ At that time, the rolling changes to pure rolling
Let the pure rolling begin after 't' seconds
• For pure rolling, we have: $\mathbf\small{|\vec{v}_{CM}|=R\,|\vec{\omega}|}$
• All points of the disc will have the same linear velocity $\mathbf\small{|\vec{v}_{CM}|}$
• So we can simply write $\mathbf\small{|\vec{v}|}$ instead of $\mathbf\small{|\vec{v}_{CM}|}$
• So we can write as:
When pure rolling begins: $\mathbf\small{|\vec{v}_{(t)}|=R\,|\vec{\omega}_{(t)}|}$
15. Now we substitute for $\mathbf\small{|\vec{v}_{(t)}|}$ and $\mathbf\small{|\vec{\omega}_{(t)}|}$
• We use the equations in (13) for the substitution. We get:
$\mathbf\small{|\vec{a}|\,t=R\,\left(|\vec{\omega}_i|-|\vec{\alpha}|\,t\right)}$
$\mathbf\small{\Rightarrow|\vec{a}|\,t=R|\vec{\omega}_i|-R|\vec{\alpha}|\,t}$
$\mathbf\small{\Rightarrow(|\vec{a}|+R|\vec{\alpha}|)\,t=R|\vec{\omega}_i|}$
• Thus we get:
Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
• Where:
♦ t is the time after which pure rolling begins
♦ R is the radius of the disc
♦ $\mathbf\small{|\vec{\omega}_i|}$ is the angular velocity at the time of touch down
♦ $\mathbf\small{|\vec{a}|=\frac{|\vec{f_k}|}{m}}$
♦ $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
We will see a solved example:
Solved example 7.46
A wheel spinning at an angular velocity of 70 rad s-1 is slowly lowered on to the ground with out any horizontal force. Radius of the wheel is 0.4 m and it's mass is 15 kg.
(a) After what time will the pure rolling begin?
(b) What is the distance traveled by the wheel before the beginning of pure rolling ?
Coefficient of kinetic friction between the wheel and the ground is 0.5
Take g = 9.8 ms-2
Solution:
Part (a):
1. We have Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
2. Calculation of $\mathbf\small{|\vec{a}|}$:
• $\mathbf\small{|\vec{a}|=\frac{|\vec{F}|}{m}=\frac{|\vec{f_k}|}{m}=\frac{\mu_k\,m\,|\vec{g}|}{m}=\mu_k|\vec{g}|}$
• So we get: $\mathbf\small{|\vec{a}|=\mu_k|\vec{g}|=(0.5 \times 9.8)}$ = 4.9 ms-2
3. Calculation of $\mathbf\small{|\vec{\alpha}|}$:
• $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
• For a disc $\mathbf\small{I=\frac{mR^2}{2}}$
• So we get: $\mathbf\small{|\vec{\alpha}|=\frac{2R\,|\vec{f_k}|}{mR^2}=\frac{2\,|\vec{f_k}|}{mR}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2\,\mu_k\,m\,|\vec{g}|}{mR}=\frac{2\,\mu_k\,|\vec{g}|}{R}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2(0.5)(9.8)}{0.4}=25.4}$ rad s-2.
4. Substituting the results from (2) and (3) in (1), we get:
$\mathbf\small{t= \frac{0.4 \times 70}{[(4.9)+0.4(25.4)]}=1.86\, \rm{s}}$
Part (b):
1. Consider the two instances:
(i) Instance at which the contact with the floor is made
(ii) Instance at which the pure rolling begins
2. From part (a), we have: Time interval between the two instances = 1.86 s
• Linear acceleration with which the wheel travels during this time interval = (0.5 × 9.8) = 4.9 ms-2.
• Initial velocity of this linear motion = 0
• So we can find the velocity at the instance when pure rolling begins
3. We can use the relation $\mathbf\small{|\vec{v}_f|=|\vec{v}_i|+|\vec{a}|\,t}$
• For our present case: $\mathbf\small{|\vec{v}_f|=0+|\vec{a}|\,t}$
$\mathbf\small{\Rightarrow|\vec{v}_f|=|\vec{a}|\,t}$
• Substituting the values, we get: $\mathbf\small{|\vec{v}_f|}$ = (4.9 × 1.86) = 9.114 ms-1.
4. Let $\mathbf\small{|\vec{x}|}$ be the distance traveled by the wheel during this interval
• We can use the relation: $\mathbf\small{|\vec{v}_f|^2-|\vec{v}_i|^2=2|\vec{a}|\,|\vec{x}|}$
• Substituting the values, we get: $\mathbf\small{9.1^2-0=2\times4.9\times|\vec{x}|}$
• Thus we get: $\mathbf\small{|\vec{x}|}$ = 8.45 m
Solved example 7.47
A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to 10 𝝅 rad s-1. Which of the two will start to roll earlier ? The co-efficient of kinetic friction is 𝝁k = 0.2
Solution:
• The solid disc and ring are placed simultaneously
• Both have the same initial angular velocity of 10 𝝅 rad s-1
• As soon as they are placed, both will move horizontally in the same direction
• But for both of them, this motion will not be 'pure rolling'
• Pure rolling will start only after some time 't' after the 'instant of touch down'
• This 't' will be different for the two objects
• We have to find which object has the smallest 't'
• Given that the radius is the same 10 cm for both the objects. So we will write:
♦ Radius of disc = Radius of ring = R = 10 cm
1. First we will consider the solid disc
We have:
Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
• Where:
♦ t is the time after which pure rolling begins
♦ R is the radius of the disc
♦ $\mathbf\small{|\vec{\omega}_i|}$ is the angular velocity at the time of touch down
♦ $\mathbf\small{|\vec{a}|=\frac{|\vec{f_k}|}{m}}$
♦ $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
2. Calculation of $\mathbf\small{|\vec{a}|}$:
• $\mathbf\small{|\vec{a}|=\frac{|\vec{F}|}{m}=\frac{|\vec{f_k}|}{m}=\frac{\mu_k\,m\,|\vec{g}|}{m}=\mu_k|\vec{g}|}$
• So we get: $\mathbf\small{|\vec{a}|=\mu_k|\vec{g}|=(0.2 \times 9.8)}$ = 1.96 ms-2
3. Calculation of $\mathbf\small{|\vec{\alpha}|}$:
• $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
• For a disc $\mathbf\small{I=\frac{mR^2}{2}}$
• So we get: $\mathbf\small{|\vec{\alpha}|=\frac{2R\,|\vec{f_k}|}{mR^2}=\frac{2\,|\vec{f_k}|}{mR}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2\,\mu_k\,m\,|\vec{g}|}{mR}=\frac{2\,\mu_k\,|\vec{g}|}{R}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{2(0.2)(9.8)}{R}=\frac{3.92}{R}}$ rad s-2.
4. Substituting the results from (2) and (3) in (1), we get:
$\mathbf\small{t_d= \frac{R \times 10 \pi}{[(1.96)+R(\frac{3.92}{R})]}=1.7\pi R\, \rm{s}}$
5. Now we consider the ring
We have:
Eq.7.35: $\mathbf\small{t= \frac{R|\vec{\omega}_i|}{(|\vec{a}|+R|\vec{\alpha}|)}}$
6. Calculation of $\mathbf\small{|\vec{a}|}$:
• $\mathbf\small{|\vec{a}|=\frac{|\vec{F}|}{m}=\frac{|\vec{f_k}|}{m}=\frac{\mu_k\,m\,|\vec{g}|}{m}=\mu_k|\vec{g}|}$
• So we get: $\mathbf\small{|\vec{a}|=\mu_k|\vec{g}|=(0.2 \times 9.8)}$ = 1.96 ms-2
7. Calculation of $\mathbf\small{|\vec{\alpha}|}$:
• $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{\tau}|}{I}=\frac{R\,|\vec{f_k}|}{I}}$
• For a ring $\mathbf\small{I=mR^2}$
• So we get: $\mathbf\small{|\vec{\alpha}|=\frac{R\,|\vec{f_k}|}{mR^2}=\frac{|\vec{f_k}|}{mR}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{\mu_k\,m\,|\vec{g}|}{mR}=\frac{\mu_k\,|\vec{g}|}{R}}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=\frac{(0.2)(9.8)}{R}=\frac{1.96}{R}}$ rad s-2.
8. Substituting the results from (6) and (7) in (5), we get:
$\mathbf\small{t_r= \frac{R \times 10 \pi}{[(1.96)+R(\frac{1.96}{R})]}=2.55\pi R\, \rm{s}}$
9. Let us write the results together:
• From (4) we have: td = 1.7 𝝅R
• From (8) we have: td = 2.55 𝝅R
• Obviously, td is lesser. That means, the disc will start pure rolling earlier
A practical application:
• We have seen that, a spinning disc, if brought into contact with the ground will slip
• But when we set a car into motion, the rear wheels are not 'already spinning'
• We slowly transmit the energy from the engine to the rear wheels
• So the rear wheels slowly begins to roll. There will not be any slipping
• If a large energy is transmitted in a small interval of time (that is 'large power'), those wheels will spin
• We want to know the maximum power that can be transmitted so that, spinning does not occur
• The following solved example demonstrates this aspect:
Solved example 7.48
A rear wheel drive car has a total mass of 1200 kg. The wheels have a radius of 0.3 m
(a) What is the maximum torque that can be imparted to the particles on the periphery of the rear wheels?
(b) What is the maximum linear acceleration with which the driver can set the car into motion?
Coefficient of static friction = 0.6
Solution:
Part (a):
1. Assume that the total mass of the car is distributed equally among the four wheels
• So mass carried by each wheel = 1200⁄4 = 300 kg
⇒ Weight acting down on each wheel = (300 × 9.8) = 2940 N
⇒ Force of friction on each wheel = 𝝁smg = (0.6 × 2940) = 1764 N
2. This is the maximum force which the ground can apply on the particle P
('Particle P' is the particle of the wheel which is in contact with the ground. See fig.7.145.b above)
• Also, this '1764 N frictional force' acts towards the right
3. This frictional force is 'made available' because, particle P applies a force tangentially towards the left
• If this 'force applied by the particle' is greater than the 'force available from the ground', then the wheel will slip
4. So we can write:
The maximum force which the particle P can exert
= 1764 N
5. For the next step, we first write some basics which we have already learned in rotational motion:
(i) The wheel is initially at rest. So the initial angular velocity = 0
(ii) We gradually apply the energy produced in the engine onto the wheel
• So the wheel gradually begins to rotate. That means it's angular velocity changes from zero
(iii) A change in angular velocity can be brought about only by an application of a torque
• When we transmit the energy from the engine to the wheel, we are indeed applying a torque
(iv) So every particle on the wheel will be experiencing a torque
• The torque experienced by a particle will depend on it's distance 'r' from the center of the wheel
• We have the equation: Torque = Force × perpendicular distance from center
6. In the present case:
• Let the torque experience by the particle P be $\mathbf\small{|\vec{\tau}|}$
• Let the tangential force experienced by the particle P be $\mathbf\small{|\vec{F}|}$
(Remember that the above torque and force are due to the energy transmitted from the engine)
• We can write: $\mathbf\small{|\vec{\tau}|=|\vec{F}|R}$
7. The particle P transmits this $\mathbf\small{|\vec{F}|}$ tangentially to the ground
• When the ground experience this $\mathbf\small{|\vec{F}|}$, it puts up an equal and opposite reaction force. This reaction force is the friction
8. We saw that the maximum friction available is 1764 N
• So $\mathbf\small{|\vec{F}|}$ must not exceed 1764 N
• If $\mathbf\small{|\vec{F}|}$ exceeds this value, the wheel will slip
9. So from (6), we get: $\mathbf\small{|\vec{\tau}|}$ = (1764 × 0.3) = 529.2 N m
• We can write:
The energy transmitted (or the power, which is 'energy transmitted per second') from the engine should be in such a way that, the torque experienced by the particles at the periphery of the wheel do not exceed 529.2 N m
• A detailed description is appropriate here:
(i) Inside the engine, chemical energy is converted to mechanical energy
• This energy is used to rotate the rear wheels of the car. Thus the car moves forward
(ii) Since the 'rotation of the wheels' is what we want, we consider the 'torque' rather than the 'force'
(iii) We know that: Work done by the force = force × linear displacement
• Similarly, Work done by torque = torque × angular displacement
(We saw this in chapter 7.29)
(iv) 'Work done' is actually energy available from the engine
(v) We see many advertisements (for engines) specifying the torque which can be produced
• Another way for specifying the capacity of the engine is by giving the value of power
• Power is the energy available in unit time (We saw this in chapter 7.30)
(vi) Whether be it power or energy, for our present discussions, we need to consider only two points:
(i) The car engine produces the energy (or power) to rotate the rear wheels of the car
(Most cars are rear wheel driven. Which means energy is given to the rear wheels only)
(ii) The car engine does not give a push or pull in the linear direction
• We have seen that, a spinning disc, if brought into contact with the ground will slip
• But when we set a car into motion, the rear wheels are not 'already spinning'
• We slowly transmit the energy from the engine to the rear wheels
• So the rear wheels slowly begins to roll. There will not be any slipping
• If a large energy is transmitted in a small interval of time (that is 'large power'), those wheels will spin
• We want to know the maximum power that can be transmitted so that, spinning does not occur
• The following solved example demonstrates this aspect:
Solved example 7.48
A rear wheel drive car has a total mass of 1200 kg. The wheels have a radius of 0.3 m
(a) What is the maximum torque that can be imparted to the particles on the periphery of the rear wheels?
(b) What is the maximum linear acceleration with which the driver can set the car into motion?
Coefficient of static friction = 0.6
Solution:
Part (a):
1. Assume that the total mass of the car is distributed equally among the four wheels
• So mass carried by each wheel = 1200⁄4 = 300 kg
⇒ Weight acting down on each wheel = (300 × 9.8) = 2940 N
⇒ Force of friction on each wheel = 𝝁smg = (0.6 × 2940) = 1764 N
2. This is the maximum force which the ground can apply on the particle P
('Particle P' is the particle of the wheel which is in contact with the ground. See fig.7.145.b above)
• Also, this '1764 N frictional force' acts towards the right
3. This frictional force is 'made available' because, particle P applies a force tangentially towards the left
• If this 'force applied by the particle' is greater than the 'force available from the ground', then the wheel will slip
4. So we can write:
The maximum force which the particle P can exert
= 1764 N
5. For the next step, we first write some basics which we have already learned in rotational motion:
(i) The wheel is initially at rest. So the initial angular velocity = 0
(ii) We gradually apply the energy produced in the engine onto the wheel
• So the wheel gradually begins to rotate. That means it's angular velocity changes from zero
(iii) A change in angular velocity can be brought about only by an application of a torque
• When we transmit the energy from the engine to the wheel, we are indeed applying a torque
(iv) So every particle on the wheel will be experiencing a torque
• The torque experienced by a particle will depend on it's distance 'r' from the center of the wheel
• We have the equation: Torque = Force × perpendicular distance from center
6. In the present case:
• Let the torque experience by the particle P be $\mathbf\small{|\vec{\tau}|}$
• Let the tangential force experienced by the particle P be $\mathbf\small{|\vec{F}|}$
(Remember that the above torque and force are due to the energy transmitted from the engine)
• We can write: $\mathbf\small{|\vec{\tau}|=|\vec{F}|R}$
7. The particle P transmits this $\mathbf\small{|\vec{F}|}$ tangentially to the ground
• When the ground experience this $\mathbf\small{|\vec{F}|}$, it puts up an equal and opposite reaction force. This reaction force is the friction
8. We saw that the maximum friction available is 1764 N
• So $\mathbf\small{|\vec{F}|}$ must not exceed 1764 N
• If $\mathbf\small{|\vec{F}|}$ exceeds this value, the wheel will slip
9. So from (6), we get: $\mathbf\small{|\vec{\tau}|}$ = (1764 × 0.3) = 529.2 N m
• We can write:
The energy transmitted (or the power, which is 'energy transmitted per second') from the engine should be in such a way that, the torque experienced by the particles at the periphery of the wheel do not exceed 529.2 N m
• We have been using this sentence many times: 'Energy produced in the engine is transmitted to the rear wheels'
(i) Inside the engine, chemical energy is converted to mechanical energy
• This energy is used to rotate the rear wheels of the car. Thus the car moves forward
(ii) Since the 'rotation of the wheels' is what we want, we consider the 'torque' rather than the 'force'
(iii) We know that: Work done by the force = force × linear displacement
• Similarly, Work done by torque = torque × angular displacement
(We saw this in chapter 7.29)
(iv) 'Work done' is actually energy available from the engine
(v) We see many advertisements (for engines) specifying the torque which can be produced
• Another way for specifying the capacity of the engine is by giving the value of power
• Power is the energy available in unit time (We saw this in chapter 7.30)
(vi) Whether be it power or energy, for our present discussions, we need to consider only two points:
(i) The car engine produces the energy (or power) to rotate the rear wheels of the car
(Most cars are rear wheel driven. Which means energy is given to the rear wheels only)
(ii) The car engine does not give a push or pull in the linear direction
Now we take up part (b):
1. Frictional force available at one rear wheel = 1764 N
• So total frictional force available at the two rear wheels = (1764 × 2) = 3528 N
[Remember that, this forward force of 3528 N is not made available (from the ground) free of cost. It is a reaction force, which can be produced only if the car tire applies the same force in the backward direction. Recall that, if a space astronaut want to move towards the left, he must push onto a massive object on his right side. It is the only option available to him unless he is wearing a jet thruster backpack]
3. Given that, total mass of the car = 1200 kg
• So maximum possible acceleration of the car = Force⁄mass = 3528⁄1200 = 2.94 m s-2
4. So the driver must set the car into motion in such a way that, the acceleration of the car does not exceed 2.94 m s-2
• What does that mean?
Let us analyze:
(i) The car is initially at rest
• When it is set in motion, it attains a velocity v
(ii) So there is a 'change in velocity' because, velocity changes from 0 to v
• If there is a change in velocity, obviously, there must be an acceleration
• So without an acceleration, no vehicle can attain a required velocity
(iii) We know that $\mathbf\small{a=\frac{v_2-v_1}{t}}$
• 't' is in the denominator
• So a large acceleration implies that, a large 'change in velocity' is brought about in a small time
• Such a large acceleration will cause the car to jerk and the rear wheels will spin
(iv) So the driver must not give a large acceleration
• In our present case, the acceleration must not exceed 2.94 ms-2
1. Frictional force available at one rear wheel = 1764 N
• So total frictional force available at the two rear wheels = (1764 × 2) = 3528 N
[Remember that, this forward force of 3528 N is not made available (from the ground) free of cost. It is a reaction force, which can be produced only if the car tire applies the same force in the backward direction. Recall that, if a space astronaut want to move towards the left, he must push onto a massive object on his right side. It is the only option available to him unless he is wearing a jet thruster backpack]
3. Given that, total mass of the car = 1200 kg
• So maximum possible acceleration of the car = Force⁄mass = 3528⁄1200 = 2.94 m s-2
4. So the driver must set the car into motion in such a way that, the acceleration of the car does not exceed 2.94 m s-2
• What does that mean?
Let us analyze:
(i) The car is initially at rest
• When it is set in motion, it attains a velocity v
(ii) So there is a 'change in velocity' because, velocity changes from 0 to v
• If there is a change in velocity, obviously, there must be an acceleration
• So without an acceleration, no vehicle can attain a required velocity
(iii) We know that $\mathbf\small{a=\frac{v_2-v_1}{t}}$
• 't' is in the denominator
• So a large acceleration implies that, a large 'change in velocity' is brought about in a small time
• Such a large acceleration will cause the car to jerk and the rear wheels will spin
(iv) So the driver must not give a large acceleration
• In our present case, the acceleration must not exceed 2.94 ms-2
Now we will derive an interesting relation
1. In the above solved example 7.48, we saw that:
• The car tires are rolling without slipping
• At the same time, the car is moving with acceleration
2. The ‘acceleration of the car’ is ‘linear acceleration’
• Since the car is a rigid body, 'every non-rotating particle' in it will be having the same linear acceleration
• So the center of the wheel will also be having the same linear acceleration
3. Now consider a disc
• It is rolling with out slipping
• At the same time, it’s center is experiencing linear acceleration
4. We have:
Eq.7.32: $\mathbf\small{|\vec{v}_{CM}|=R|\vec{\omega}|}$
5. Since our present disc is moving with linear acceleration, the linear velocity will change
• So the right side of Eq.7.32 will also change
• But R is a constant. So we can say, if the disc is in pure rolling and is moving with linear acceleration, the angular $\mathbf\small{|\vec{\omega}|}$ also changes
• If the angular velocity changes, it means that there is angular acceleration
6. Consider the instant at which the reading in the stop watch is 0
• Let the magnitude of linear velocity be $\mathbf\small{|\vec{v}_0|}$
• Let the magnitude of angular velocity be $\mathbf\small{|\vec{\omega}_0|}$
7. Consider the instant at which the reading in the stop watch is t
• Let the magnitude of linear velocity be $\mathbf\small{|\vec{v}_t|}$
• Let the magnitude of angular velocity be $\mathbf\small{|\vec{\omega}_t|}$
8. For linear motion, we have: $\mathbf\small{|\vec{v}_t|=|\vec{v}_0|+|\vec{a}|\,t}$
• So we can write: $\mathbf\small{t=\frac{|\vec{v}_t|-|\vec{v}_0|}{|\vec{a}|}}$
9. For angular motion, we have: $\mathbf\small{|\vec{\omega}_t|=|\vec{\omega}_0|+|\vec{\alpha}|\,t}$
• So we can write: $\mathbf\small{t=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
10. The 't' is same in both (7) and (8)
• So we can equate them. We get: $\mathbf\small{\frac{|\vec{v}_t|-|\vec{v}_0|}{|\vec{a}|}=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
11. $\mathbf\small{|\vec{\omega}_t|}$ is the angular velocity at the instant when the reading in the stopwatch is 't'
• At that instant, the linear velocity of CM is $\mathbf\small{|\vec{v}_t|}$
• Since there is no slipping, we have: $\mathbf\small{|\vec{v}_t|=R|\vec{\omega}_t|}$
12. $\mathbf\small{|\vec{\omega}_0|}$ is the angular velocity at the instant when the reading in the stopwatch is 0
• At that instant, the linear velocity of CM is $\mathbf\small{|\vec{v}_0|}$
• Since there is no slipping, we have: $\mathbf\small{|\vec{v}_0|=R|\vec{\omega}_0|}$
13. So the result in (9) will become:
$\mathbf\small{\frac{R|\vec{\omega}_t|-R|\vec{\omega}_0|}{|\vec{a}|}=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
$\mathbf\small{\Rightarrow\frac{R(|\vec{\omega}_t|-|\vec{\omega}_0|)}{|\vec{a}|}=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
$\mathbf\small{\Rightarrow\frac{R}{|\vec{a}|}=\frac{1}{|\vec{\alpha}|}}$
Thus we get:
Eq.7.36: $\mathbf\small{|\vec{a}|=R\,|\vec{\alpha}|}$
14. In the above solved example 7.48, we have: $\mathbf\small{|\vec{a}_{max}|}$ = 2.94 ms-2
• That means, the maximum allowable linear acceleration for the center of the wheel is 2.94 ms-2
• The wheel is to roll with out slipping. So the relation $\mathbf\small{|\vec{a}|=R\,|\vec{\alpha}|}$ is valid
• We can write:$\mathbf\small{2.94=0.3\,|\vec{\alpha}|}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=9.8}$ rad s-2
• That means, the driver must accelerate the car in such a way that, the angular acceleration of the rear wheels do not exceed 9.8 rad s-2
1. In the above solved example 7.48, we saw that:
• The car tires are rolling without slipping
• At the same time, the car is moving with acceleration
2. The ‘acceleration of the car’ is ‘linear acceleration’
• Since the car is a rigid body, 'every non-rotating particle' in it will be having the same linear acceleration
• So the center of the wheel will also be having the same linear acceleration
3. Now consider a disc
• It is rolling with out slipping
• At the same time, it’s center is experiencing linear acceleration
4. We have:
Eq.7.32: $\mathbf\small{|\vec{v}_{CM}|=R|\vec{\omega}|}$
5. Since our present disc is moving with linear acceleration, the linear velocity will change
• So the right side of Eq.7.32 will also change
• But R is a constant. So we can say, if the disc is in pure rolling and is moving with linear acceleration, the angular $\mathbf\small{|\vec{\omega}|}$ also changes
• If the angular velocity changes, it means that there is angular acceleration
6. Consider the instant at which the reading in the stop watch is 0
• Let the magnitude of linear velocity be $\mathbf\small{|\vec{v}_0|}$
• Let the magnitude of angular velocity be $\mathbf\small{|\vec{\omega}_0|}$
7. Consider the instant at which the reading in the stop watch is t
• Let the magnitude of linear velocity be $\mathbf\small{|\vec{v}_t|}$
• Let the magnitude of angular velocity be $\mathbf\small{|\vec{\omega}_t|}$
8. For linear motion, we have: $\mathbf\small{|\vec{v}_t|=|\vec{v}_0|+|\vec{a}|\,t}$
• So we can write: $\mathbf\small{t=\frac{|\vec{v}_t|-|\vec{v}_0|}{|\vec{a}|}}$
9. For angular motion, we have: $\mathbf\small{|\vec{\omega}_t|=|\vec{\omega}_0|+|\vec{\alpha}|\,t}$
• So we can write: $\mathbf\small{t=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
10. The 't' is same in both (7) and (8)
• So we can equate them. We get: $\mathbf\small{\frac{|\vec{v}_t|-|\vec{v}_0|}{|\vec{a}|}=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
11. $\mathbf\small{|\vec{\omega}_t|}$ is the angular velocity at the instant when the reading in the stopwatch is 't'
• At that instant, the linear velocity of CM is $\mathbf\small{|\vec{v}_t|}$
• Since there is no slipping, we have: $\mathbf\small{|\vec{v}_t|=R|\vec{\omega}_t|}$
12. $\mathbf\small{|\vec{\omega}_0|}$ is the angular velocity at the instant when the reading in the stopwatch is 0
• At that instant, the linear velocity of CM is $\mathbf\small{|\vec{v}_0|}$
• Since there is no slipping, we have: $\mathbf\small{|\vec{v}_0|=R|\vec{\omega}_0|}$
13. So the result in (9) will become:
$\mathbf\small{\frac{R|\vec{\omega}_t|-R|\vec{\omega}_0|}{|\vec{a}|}=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
$\mathbf\small{\Rightarrow\frac{R(|\vec{\omega}_t|-|\vec{\omega}_0|)}{|\vec{a}|}=\frac{|\vec{\omega}_t|-|\vec{\omega}_0|}{|\vec{\alpha}|}}$
$\mathbf\small{\Rightarrow\frac{R}{|\vec{a}|}=\frac{1}{|\vec{\alpha}|}}$
Thus we get:
Eq.7.36: $\mathbf\small{|\vec{a}|=R\,|\vec{\alpha}|}$
14. In the above solved example 7.48, we have: $\mathbf\small{|\vec{a}_{max}|}$ = 2.94 ms-2
• That means, the maximum allowable linear acceleration for the center of the wheel is 2.94 ms-2
• The wheel is to roll with out slipping. So the relation $\mathbf\small{|\vec{a}|=R\,|\vec{\alpha}|}$ is valid
• We can write:$\mathbf\small{2.94=0.3\,|\vec{\alpha}|}$
$\mathbf\small{\Rightarrow|\vec{\alpha}|=9.8}$ rad s-2
• That means, the driver must accelerate the car in such a way that, the angular acceleration of the rear wheels do not exceed 9.8 rad s-2
So we have completed a discussion on the rotation of car wheels which are able to self propagate. In the next section, we will see the case of carts which need external force to propagate
No comments:
Post a Comment