Chapter 5.23 - Solved examples on Centripetal Force

In the previous section, we saw centripetal force. In this section, we will see some solved examples.

Solved example 5.40
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ?
Solution:
Part (a):
1. Speed of the stone = 40 rev./min = $\mathbf\small{\frac{40}{60}}$ rev./sec = $\mathbf\small{\frac{2}{3}}$ rev./sec
• That means, the stone completes 'two third of a revolution' in one second
• That means, the stone turns through $\mathbf\small{(\frac{2}{3}\times 2 \pi)=\frac{4 \pi}{3}}$ radians in one second
• Thus we get: Angular velocity of the stone $\mathbf\small{\omega=\frac{4 \pi}{3}}$ rad/sec
2. Centripetal force required = $\mathbf\small{|\vec{f_c}|=\frac{m|\vec{v}|^2}{R}=\frac{mR^2 \omega^2}{R}=mR\omega^2}$
($\mathbf\small{\because |\vec{v}|=R\omega}$)
Substituting the values, we get: $\mathbf\small{|\vec{f_c}|=0.25 \times 1.5 \times \left\lgroup\frac{4 \pi}{3}\right\rgroup ^2=6.57\, \text{N}}$
3. The required centripetal force for the stone is provided by the tension in the string. 
Thus we get: Tension in the string = 6.57 N
Part (b):
1. Let the maximum allowable speed be $\mathbf\small{\omega_{max}}$ 
• Then we can write: $\mathbf\small{|\vec{f_c}|=200 \, \text{N}=mR\omega_{max}^2}$
• Substituting the values, we get: $\mathbf\small{200 =0.25 \times 1.5 \times\omega_{max}^2}$
• Thus we get: $\mathbf\small{\omega_{max}}$ = 23.094 rad/sec
2. So the stone can cover 23.094 radians in 1 second
• Then number of revolutions in 1 second = $\mathbf\small{\frac{23.094}{2\pi}=3.68}$
• We can express the maximum allowable speed in terms of linear velocity also. We have:
$\mathbf\small{|\vec{v_{max}}|=R\omega_{max}=1.5\times23.094=34.64\,\text{ms}^{-1}}$

Solved example 5.41
A disc revolves with a speed of 33¹⁄₃ rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record ?
Solution:
The arrangement is shown in fig.5.92(a) below:

Fig.5.92

1. Angular speed of the disc = 33¹⁄₃ rev./min = ¹⁰⁰⁄₃ rev./min = $\mathbf\small{\frac{100}{180}}$ rev./sec = $\mathbf\small{\frac{5}{9}}$ rev./sec
• That means, the disc turns through $\mathbf\small{(\frac{5}{9}\times 2 \pi)=\frac{10 \pi}{9}}$ radians in one second
• Thus we get: Angular speed of the disc $\mathbf\small{\omega=\frac{10 \pi}{9}}$ rad/sec
2. Though the angular speed of the two coins are the same, their linear speeds will be different (Details here)
• Linear speed of coin 1 = $\mathbf\small{|\vec{v_{c1}}|=R_1\omega=0.04\times\frac{10 \pi}{9}=\frac{0.4 \pi}{9}}$
• Linear speed of coin 2 = $\mathbf\small{|\vec{v_{c2}}|=R_2\omega=0.14\times\frac{10 \pi}{9}=\frac{1.4 \pi}{9}}$
3. Centripetal force required for the first coin =
$\mathbf\small{|\vec{f_{c1}}|=\frac{m|\vec{v_1}|^2}{R_1}=m \times \frac{1}{0.04} \times \left\lgroup\frac{0.4 \pi}{9}\right\rgroup ^2=\frac{4\pi^2m}{81}=0.487m\, \text{N}}$ 
• Centripetal force required for the second coin =
$\mathbf\small{|\vec{f_{c2}}|=\frac{m|\vec{v_2}|^2}{R_2}=m \times \frac{1}{0.14} \times \left\lgroup\frac{1.4 \pi}{9}\right\rgroup ^2=\frac{14\pi^2m}{81}=1.706m\, \text{N}}$
4. For both the coins, the centripetal force will be provided by the friction
Magnitude of the frictional force for each coin = $\mathbf\small{\mu_smg=0.15 \times m \times 10=1.5m \, \text{N}}$
• Comparing with the results in (3), we see that:
(i) The centripetal force required for the inner coin is less than the force available from friction
(ii) The centripetal force required for the outer coin is more than the force available from friction
■ So we can write: 
• The inner coin will revolve with the disc
• The outer coin will slip away

Solved example 5.42
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. See fig.5.92(b) above. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed ?
Solution:
1. When the man stands inside a stationary hollow cylinder, his weight is supported by the floor of the cylinder
• When the cylinder rotates, he will be pressed against the inner surface of the cylinder. This pressing is due to the centripetal force
• Because of this pressing, frictional force will develop between his clothes and the wall of the cylinder
• Our first step is to find the magnitude of this frictional force
We have:
Static frictional force = $\mathbf\small{|\vec{f_s}|=\mu_s|\vec{F_N}|=0.15 \times |\vec{F_N}|}$
2. But $\mathbf\small{|\vec{F_N}|}$ =Centripetal force = $\mathbf\small{|\vec{f_c}|=mR\omega_{min}^2}$
• Where $\mathbf\small{\omega_{min}}$ is the minimum angular velocity required
• Substituting the known values, we get: $\mathbf\small{|\vec{f_c}|=m \times 3 \times \left\lgroup\omega_{min}\right\rgroup ^2}$
3. We can put this value of $\mathbf\small{|\vec{f_c}|}$ in the place of $\mathbf\small{|\vec{F_N}|}$ in (1). We get:
$\mathbf\small{|\vec{f_s}|=0.15 \times m \times 3 \times \left\lgroup\omega_{min}\right\rgroup ^2}$
4. The above result in (3) is the static friction which holds the man from falling down when the floor is removed
• So this friction must be equal to 'mg'. We can write:
$\mathbf\small{0.15 \times m \times 3 \times \left\lgroup\omega_{min}\right\rgroup ^2=mg}$
$\mathbf\small{\Longrightarrow \omega_{min}= \sqrt{\frac{g}{0.15 \times 3}}=4.71\,\text{rad/sec}}$

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In the next chapter, we will see work, energy and power

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