Question

A wheel having a radius of 10 cm is coupled by a belt to another wheel of radius 30 cm. 1st wheel increases its angular speed from rest at a uniform rate of $1.57\text{ }rad/{{s}^{2}}$. The time for the 2nd wheel to reach a rotational speed of $100\text{ }rev/min$ is: (assume that the belt does not slip)

Answer 1

**Hint : **To find the time taken by the second wheel to reach $100\text{ }rev/\min$ we will use the formula of kinematics $v=u+at$ but in terms of angular velocity and angular acceleration as $\omega ={{\omega }_{o}}+\alpha t$. For that we need to find the relationship between linear velocity and acceleration to angular velocity and acceleration i.e.:
$v=r\omega \ and a=r\alpha$
where $v$ is the linear velocity, r is the radius of rotation, $\omega$ and $\alpha$ are angular velocity and acceleration respectively, ${{\omega }_{o}}$ is the initial angular velocity.

Complete step by step solution:
First let us change the angular revolution from $rev/\min$ to $rad/{{s}^{2}}$ by multiplying the number by $\dfrac{2\pi }{60}$ as shown below:
The angular velocity is $100\times \dfrac{2\pi }{{{60}^{\circ }}}=\dfrac{10\pi }{3}rad/s$
Now as we know the angular velocity of the second wheel we find the angular acceleration of the second wheel as well where we equate both the linear acceleration together, the linear acceleration of both the wheels are same as the linear acceleration doesn’t depends upon the radius of the wheel as here the linear acceleration is that of the belt passing over the wheel hence if ${{r}_{1}},{{r}_{2}}$ are radius $10cm$ and $30cm$ then, the angular acceleration of the second wheel is:
Linear acceleration of the first wheel is ${{a}_{first}}=r\alpha$
$\Rightarrow {{a}_{first}}=10\times 1.57$
Linear acceleration of the first wheel is ${{a}_{\sec ond}}=r\alpha$
$\Rightarrow {{a}_{\sec ond}}=30\times {{\alpha }_{\sec ond}}$
Equating them we get:
$\Rightarrow 10\times 1.57=30\times {{\alpha }_{\sec ond}}$
$\Rightarrow {{\alpha }_{\sec ond}}=\dfrac{10\times 1.57}{30}$
$\Rightarrow {{\alpha }_{\sec ond}}=0.52rad/{{\sec }^{2}}$
With angular acceleration found we use the kinematic formula to find the time required for the rotation:
$\omega ={{\omega }_{o}}+\alpha t$
With the final angular velocity as $\omega =\dfrac{10\pi }{3}$ and ${{\omega }_{o}}=0$ whereas the angular acceleration of the second wheel is taken as $0.52rad/{{\sec }^{2}}$.
$\Rightarrow t=\dfrac{\omega }{\alpha }$
$\Rightarrow t=\dfrac{\dfrac{10}{3}\pi }{0.52}$
$\Rightarrow t=20.12\approx 20\sec$

**Note : **The linear acceleration is equal for both the wheels because there is no slippage due to slippage both the velocity and acceleration would have been different. Hence, the no slippage part tells us that we have to equate to find the unknown acceleration and the initial velocity is zero for the second wheel as the second wheel is moving due to the motion of the first wheel.