Question

A car is moving at a speed of 72km/hr. the diameter of its wheel is 0.5m. If the wheels are stopped in 20 rotations by applying the brakes, then angular retardation produced by the brakes is
$\begin{aligned} & a)-25.5rad/{{s}^{2}} \\\ & b)-29.5rad/{{s}^{2}} \\\ & c)-33.5rad/{{s}^{2}} \\\ & d)-45.5rad/{{s}^{2}} \\\ \end{aligned}$

Answer 1

It is given in the question that the car travels with the speed of 72km/hr. We need to express this in terms of angular velocity of the wheels of the car. It is given that the retardation produced by the brakes stops the car after 20 rotations. Hence we can determine the angular retardation by using equations of rotational dynamics.
Formula used:
$v=r\omega$
$\omega ={{\omega }_{\circ }}+\alpha t$
$\theta ={{\omega }_{\circ }}t+\dfrac{\alpha {{t}^{2}}}{2}$

Complete answer:
The linear velocity of the wheel i.e. (v)is given by $v=r\omega$ where r is the radius of the wheel and $\omega$is the angular velocity of the wheel. The diameter of the wheel of the car is 0.5m and hence it’s radius (0.5/2) m. The linear velocity of the car is given as 72km/hr, hence using the above equation the angular velocity of the car is given by,
$\begin{aligned} & v=r\omega \\\ & \Rightarrow 72kms/hr=\dfrac{0.5}{2}\omega \\\ & \Rightarrow \dfrac{72\times {{10}^{3}}}{60\times 60}m{{s}^{-1}}=\dfrac{0.5}{2}\omega \\\ & \Rightarrow 20m{{s}^{-1}}=\dfrac{0.5}{2}\omega \\\ & \Rightarrow \omega =\dfrac{20\times 2}{0.5}=\dfrac{200\times 2}{5}=80rad/s \\\ \end{aligned}$
The above angular velocity is the initial angular velocity of the car. Now let us introduce ourselves to the two rotational dynamics equations.
$\omega ={{\omega }_{\circ }}+\alpha t.....(1)$ where $\omega$ is the final angular velocity of the of the body under rotational motion, ${{\omega }_{\circ }}$ is the initial angular velocity of the of the body under rotational motion $\alpha$ is the angular acceleration or retardation of the of the body from its initial to the final angular velocity in time t. Now let us define the angular displacement of a body performing rotational motion.
$\theta ={{\omega }_{\circ }}t+\dfrac{\alpha {{t}^{2}}}{2}...(2)$ where $\theta$ is the angular displacement of the body. The car is finally brought to rest by applying the brakes. Therefore its final angular velocity will be zero. The initial angular velocity of the car we found as 80rad/sec. hence using equation 1 the angular retardation is,
$\begin{aligned} & \omega ={{\omega }_{\circ }}+\alpha t \\\ & 0=80+\alpha t \\\ & \alpha =-\dfrac{80}{t}...(3) \\\ \end{aligned}$
It is given that the car is stopped after 20 rotations. Hence the angular distance covered by the car is,
$\begin{aligned} & \theta ={{\omega }_{\circ }}t+\dfrac{\alpha {{t}^{2}}}{2},\text{since 1 rotation is equal to 2 }\\!\\!\pi\\!\\!\text{ } \\\ & \Rightarrow 20\times 2\pi =80t+\dfrac{\alpha {{t}^{2}}}{2}\text{, since }\alpha =-\dfrac{80}{t} \\\ & \Rightarrow 20\times 2\pi =80t-\dfrac{80{{t}^{2}}}{2t} \\\ & \Rightarrow 20\times 2\times 3.14=80t-40t \\\ & \Rightarrow 125.6=40t \\\ & \Rightarrow t=3.14\sec \\\ \end{aligned}$
Substituting the time for which the wheels face retardation in equation 3 we get,
$\begin{aligned} & \alpha =-\dfrac{80}{t} \\\ & \Rightarrow \alpha =-\dfrac{80rad/s}{3.14s}=-25.47rad/{{s}^{2}} \\\ & \Rightarrow \alpha =-25.5rad/{{s}^{2}} \\\ \end{aligned}$

Therefore the correct answer of the above question is option a.

Note:
The minus sign next to the angular acceleration indicates retardation. Since force on a body is proportional to linear acceleration, the minus sign indicates the force acts in the opposite direction of motion. Therefore eventually the body comes to rest.