In the previous section, we saw the maximum acceleration with which a car can be set into motion. In this section we will see the case of a cart.
• We have seen that, a cart needs external force to move. If this force is too large, the wheels will slip
• So there is a maximum allowable force. Our next aim is to find the magnitude of this allowable force
1. Consider the wheel shown in fig.7.146(a) below
It has a mass of 7 kg. It's radius is 0.35 m
𝝁s = 0.8 and 𝝁k = 0.6
2. A force is being applied at it's center
• This force is denoted as $\mathbf\small{|\vec{F}_{max}|}$
• This is because, it is the maximum force allowable. If a greater force is applied, the wheel will slide instead of rolling
3. Fig.b shows the remaining forces also which are acting on the wheel
• The weight $\mathbf\small{m|\vec{g}|}$ acts downwards at the center
• The normal force $\mathbf\small{|\vec{N}|}$ acts upwards from the point of contact with the ground
• We have the frictional force $\mathbf\small{|\vec{f_s}|=\mu_s|\vec{N}|=\mu_sm|\vec{g}|}$
• We use the static friction because, we want the wheel to be in 'pure rolling' motion
• Also we know that, for a cart wheel, the frictional force is opposite to the direction of motion
4. This frictional force is the 'cause of the torque' which rotates the wheel
• Because, it is the only force which does not pass through the center of the wheel
5. Now we will do the further calculations as two sets
• Set 1 is related to rotational motion
• Set 2 is related to translational motion
6. Set 1:
(i) The frictional force is the 'cause of the torque'
• So we have: $\mathbf\small{|\vec{\tau}|=|\vec{f_s}|R=\mu_sm|\vec{g}|R}$
• $\mathbf\small{\mu_sm|\vec{g}|R}$ is the maximum frictional force possible
• So we can write: $\mathbf\small{|\vec{\tau}_{max}|=\mu_sm|\vec{g}|R}$
(ii) But $\mathbf\small{|\vec{\tau}|=I\,|\vec{\alpha}|}$
• Where:
♦ $\mathbf\small{I}$ is the moment of inertia of the wheel
♦ $\mathbf\small{|\vec{\alpha}|}$ is the angular acceleration of the wheel
(iii) For pure rolling, we have eq.7.36 that we derived in the previous section: $\mathbf\small{|\vec{a}|=R\,|\vec{\alpha}|}$
• Rearranging this, we get: $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{a}|}{R}}$
(iv) For our present case, we must consider the maximum allowable angular acceleration
• So we must denote it as $\mathbf\small{|\vec{\alpha}_{max}|}$
• So we have: $\mathbf\small{|\vec{\alpha}_{max}|=\frac{|\vec{a}_{max}|}{R}}$
(v) Thus from (ii), we get: $\mathbf\small{|\vec{\tau}_{max}|=I\,\frac{|\vec{a}_{max}|}{R}}$
• Now, assuming the I of the wheel to be the same as that of the disc, we have: $\mathbf\small{I=\frac{mR^2}{2}}$
• So we get: $\mathbf\small{|\vec{\tau}_{max}|=\left(\frac{mR^2}{2}\right)\,\frac{|\vec{a}_{max}|}{R}=\frac{mR|\vec{a}_{max}|}{2}}$
(vi) So the result in (i) becomes: $\mathbf\small{\frac{mR|\vec{a}_{max}|}{2}=\mu_sm|\vec{g}|R}$
• From this we get: $\mathbf\small{|\vec{a}_{max}|= 2\mu_s|\vec{g}|}$
7. Set 2:
We have: Net force = mass × acceleration
• In our present case, Net force = $\mathbf\small{|\vec{F}_{max}|-|\vec{f_s}|}$
⇒ Net force = $\mathbf\small{|\vec{F}_{max}|-\mu_s\,m|\vec{g}|}$
8. The two sets are complete
• Now we take the result from set 1 (vi)
• And apply it in the result from set 2
• We get:
Net force = $\mathbf\small{|\vec{F}_{max}|-\mu_s\,m|\vec{g}|=m\times2\mu_s|\vec{g}|}$
• Thus we get:
Eq.7.37: $\mathbf\small{|\vec{F}_{max}|=m\times2\mu_s|\vec{g}|+\mu_s\,m|\vec{g}|=3\mu_sm|\vec{g}|}$
Substituting the values, we get: $\mathbf\small{|\vec{F}_{max}|}$ = 3 × 0.8 × 7 × 9.8 = 164.64 N
Now we will see a solved example
Solved example 7.49
The wheel shown in fig.7.146(c) above, has a radius of 0.35 m and a mass of 6 kg. A horizontal force of 85 N is applied at it's center. What is the linear acceleration of the wheel? 𝝁s = 0.4 and 𝝁k = 0.3
Take g = 9.8 ms-2
Solution:
1. First of all, we have to determine whether the wheel would slip or not, when the 85 N is applied
• For that, we find the maximum allowable force $\mathbf\small{|\vec{F}_{max}|}$
• If 85 N is greater than this maximum allowable force, the wheel will slip
2. We have Eq.7.37: $\mathbf\small{|\vec{F}_{max}|=m\times2\mu_s|\vec{g}|+\mu_s\,m|\vec{g}|=3\mu_sm|\vec{g}|}$
• Substituting the values, we get: $\mathbf\small{|\vec{F}_{max}|}$ = 3 × 0.4 × 6 × 9.8 = 70.56 N
3. So the applied force of 85 N is greater than the maximum allowable force. So the wheel will slip
4. If the wheel slips, static friction is not valid any more. We must consider only kinetic friction
• The kinetic frictional force acting on the wheel = $\mathbf\small{|\vec{f_s}|=\mu_km|\vec{g}|}$
• Substituting the values, we get:
Kinetic friction acting on the wheel = 0.3 × 6 × 9.8 = 17.64 N
5. So the net force acting on the wheel = (85 - 17.64) = 67.36 N
6. We have: Net force = mass × acceleration
• Thus we get: Acceleration of the wheel = Net force⁄mass = 67.36⁄6 = 11.23 ms-2.
Solved example 7.50
A horizontal force F acts at the center of a sphere of mass m. The coefficient of static friction between the sphere and the ground is 𝝁. What is the maximum allowable value of F so that, the sphere rolls without slipping ?
Solution:
1. Let f be the force of friction. it acts opposite to the direction of motion
• So the net force acting on the sphere = F-f
2. So the acceleration of the sphere = $\mathbf\small{a=\frac{F-f}{m}}$
3. For pure rolling, we have: $\mathbf\small{a=R\alpha}$
4. But $\mathbf\small{\alpha=\frac{\tau}{I}}$
$\mathbf\small{\Rightarrow\alpha=\frac{fR}{\frac{2mR^2}{5}}=\frac{5f}{2mR}}$
5. So from (3) we get: $\mathbf\small{a=r\alpha=R\times\frac{5f}{2mR}=\frac{5f}{2m}}$
6. So from (2), we get: $\mathbf\small{a=\frac{F-f}{m}=\frac{5f}{2m}}$
$\mathbf\small{\Rightarrow{F-f}=\frac{5f}{2}}$
$\mathbf\small{\Rightarrow F=\frac{7f}{2}}$
7. But f = 𝝁mg
So the result in (6) becomes: $\mathbf\small{F=\frac{7\mu\,mg}{2}}$
8. Thus we can write:
The applied force must be less than or equal to $\mathbf\small{\frac{7\mu\,mg}{2}}$
• We have seen that, a cart needs external force to move. If this force is too large, the wheels will slip
• So there is a maximum allowable force. Our next aim is to find the magnitude of this allowable force
1. Consider the wheel shown in fig.7.146(a) below
It has a mass of 7 kg. It's radius is 0.35 m
 |
| Fig.7.146 |
2. A force is being applied at it's center
• This force is denoted as $\mathbf\small{|\vec{F}_{max}|}$
• This is because, it is the maximum force allowable. If a greater force is applied, the wheel will slide instead of rolling
3. Fig.b shows the remaining forces also which are acting on the wheel
• The weight $\mathbf\small{m|\vec{g}|}$ acts downwards at the center
• The normal force $\mathbf\small{|\vec{N}|}$ acts upwards from the point of contact with the ground
• We have the frictional force $\mathbf\small{|\vec{f_s}|=\mu_s|\vec{N}|=\mu_sm|\vec{g}|}$
• We use the static friction because, we want the wheel to be in 'pure rolling' motion
• Also we know that, for a cart wheel, the frictional force is opposite to the direction of motion
4. This frictional force is the 'cause of the torque' which rotates the wheel
• Because, it is the only force which does not pass through the center of the wheel
5. Now we will do the further calculations as two sets
• Set 1 is related to rotational motion
• Set 2 is related to translational motion
6. Set 1:
(i) The frictional force is the 'cause of the torque'
• So we have: $\mathbf\small{|\vec{\tau}|=|\vec{f_s}|R=\mu_sm|\vec{g}|R}$
• $\mathbf\small{\mu_sm|\vec{g}|R}$ is the maximum frictional force possible
• So we can write: $\mathbf\small{|\vec{\tau}_{max}|=\mu_sm|\vec{g}|R}$
(ii) But $\mathbf\small{|\vec{\tau}|=I\,|\vec{\alpha}|}$
• Where:
♦ $\mathbf\small{I}$ is the moment of inertia of the wheel
♦ $\mathbf\small{|\vec{\alpha}|}$ is the angular acceleration of the wheel
(iii) For pure rolling, we have eq.7.36 that we derived in the previous section: $\mathbf\small{|\vec{a}|=R\,|\vec{\alpha}|}$
• Rearranging this, we get: $\mathbf\small{|\vec{\alpha}|=\frac{|\vec{a}|}{R}}$
(iv) For our present case, we must consider the maximum allowable angular acceleration
• So we must denote it as $\mathbf\small{|\vec{\alpha}_{max}|}$
• So we have: $\mathbf\small{|\vec{\alpha}_{max}|=\frac{|\vec{a}_{max}|}{R}}$
(v) Thus from (ii), we get: $\mathbf\small{|\vec{\tau}_{max}|=I\,\frac{|\vec{a}_{max}|}{R}}$
• Now, assuming the I of the wheel to be the same as that of the disc, we have: $\mathbf\small{I=\frac{mR^2}{2}}$
• So we get: $\mathbf\small{|\vec{\tau}_{max}|=\left(\frac{mR^2}{2}\right)\,\frac{|\vec{a}_{max}|}{R}=\frac{mR|\vec{a}_{max}|}{2}}$
(vi) So the result in (i) becomes: $\mathbf\small{\frac{mR|\vec{a}_{max}|}{2}=\mu_sm|\vec{g}|R}$
• From this we get: $\mathbf\small{|\vec{a}_{max}|= 2\mu_s|\vec{g}|}$
7. Set 2:
We have: Net force = mass × acceleration
• In our present case, Net force = $\mathbf\small{|\vec{F}_{max}|-|\vec{f_s}|}$
⇒ Net force = $\mathbf\small{|\vec{F}_{max}|-\mu_s\,m|\vec{g}|}$
8. The two sets are complete
• Now we take the result from set 1 (vi)
• And apply it in the result from set 2
• We get:
Net force = $\mathbf\small{|\vec{F}_{max}|-\mu_s\,m|\vec{g}|=m\times2\mu_s|\vec{g}|}$
• Thus we get:
Eq.7.37: $\mathbf\small{|\vec{F}_{max}|=m\times2\mu_s|\vec{g}|+\mu_s\,m|\vec{g}|=3\mu_sm|\vec{g}|}$
Substituting the values, we get: $\mathbf\small{|\vec{F}_{max}|}$ = 3 × 0.8 × 7 × 9.8 = 164.64 N
Now we will see a solved example
Solved example 7.49
The wheel shown in fig.7.146(c) above, has a radius of 0.35 m and a mass of 6 kg. A horizontal force of 85 N is applied at it's center. What is the linear acceleration of the wheel? 𝝁s = 0.4 and 𝝁k = 0.3
Take g = 9.8 ms-2
Solution:
1. First of all, we have to determine whether the wheel would slip or not, when the 85 N is applied
• For that, we find the maximum allowable force $\mathbf\small{|\vec{F}_{max}|}$
• If 85 N is greater than this maximum allowable force, the wheel will slip
2. We have Eq.7.37: $\mathbf\small{|\vec{F}_{max}|=m\times2\mu_s|\vec{g}|+\mu_s\,m|\vec{g}|=3\mu_sm|\vec{g}|}$
• Substituting the values, we get: $\mathbf\small{|\vec{F}_{max}|}$ = 3 × 0.4 × 6 × 9.8 = 70.56 N
3. So the applied force of 85 N is greater than the maximum allowable force. So the wheel will slip
4. If the wheel slips, static friction is not valid any more. We must consider only kinetic friction
• The kinetic frictional force acting on the wheel = $\mathbf\small{|\vec{f_s}|=\mu_km|\vec{g}|}$
• Substituting the values, we get:
Kinetic friction acting on the wheel = 0.3 × 6 × 9.8 = 17.64 N
5. So the net force acting on the wheel = (85 - 17.64) = 67.36 N
6. We have: Net force = mass × acceleration
• Thus we get: Acceleration of the wheel = Net force⁄mass = 67.36⁄6 = 11.23 ms-2.
Solved example 7.50
A horizontal force F acts at the center of a sphere of mass m. The coefficient of static friction between the sphere and the ground is 𝝁. What is the maximum allowable value of F so that, the sphere rolls without slipping ?
Solution:
1. Let f be the force of friction. it acts opposite to the direction of motion
• So the net force acting on the sphere = F-f
2. So the acceleration of the sphere = $\mathbf\small{a=\frac{F-f}{m}}$
3. For pure rolling, we have: $\mathbf\small{a=R\alpha}$
4. But $\mathbf\small{\alpha=\frac{\tau}{I}}$
$\mathbf\small{\Rightarrow\alpha=\frac{fR}{\frac{2mR^2}{5}}=\frac{5f}{2mR}}$
5. So from (3) we get: $\mathbf\small{a=r\alpha=R\times\frac{5f}{2mR}=\frac{5f}{2m}}$
6. So from (2), we get: $\mathbf\small{a=\frac{F-f}{m}=\frac{5f}{2m}}$
$\mathbf\small{\Rightarrow{F-f}=\frac{5f}{2}}$
$\mathbf\small{\Rightarrow F=\frac{7f}{2}}$
7. But f = 𝝁mg
So the result in (6) becomes: $\mathbf\small{F=\frac{7\mu\,mg}{2}}$
8. Thus we can write:
The applied force must be less than or equal to $\mathbf\small{\frac{7\mu\,mg}{2}}$
So we have completed a discussion on the rotation of cart wheels which needs external force to propagate. Analysis of such wheels are applicable to 'front wheels of cars' also. In the next section, we will see the case of rolling along inclined planes
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