Question

A small car took off a ramp at a speed of 30m/s. Immediately after leaving the ramp, the driver applied brakes on all the wheels. The brakes retarded the wheels uniformly to bring them to rest in 2 second. Calculate the angle by which the car will rotate about its centre of mass in the 2 second interval after leaving the ramp. Radius of each wheel is $r = 0.30m$. Moment of inertia of the car along with the driver, about the relevant axis through its centre of mass is $I_M = 80kgm^2$ and the moment of inertia of each pair of wheels about their respective axles is $0.3kgm^2$. Assume that the car remained in air for more than 2 second. Also assume that before take-off the wheels rolled without sliding.

Answer 1

0.75 radians

Answer 2

To solve this problem, we need to determine the angular acceleration of the car body due to the braking action and then use kinematic equations to find the total angle of rotation.

1. Calculate the initial angular velocity of the wheels:

Before take-off, the wheels rolled without sliding. This means the linear speed of the car is related to the angular speed of its wheels by the formula $v = \omega_0 r$.

Given: Linear speed of the car, $v = 30 \text{ m/s}$ Radius of each wheel, $r = 0.30 \text{ m}$

Initial angular velocity of the wheels: $\omega_0 = \frac{v}{r} = \frac{30 \text{ m/s}}{0.30 \text{ m}} = 100 \text{ rad/s}$

2. Calculate the angular deceleration of the wheels:

The brakes bring the wheels to rest uniformly in 2 seconds.

Given: Initial angular velocity, $\omega_0 = 100 \text{ rad/s}$ Final angular velocity, $\omega_f = 0 \text{ rad/s}$ Time, $t = 2 \text{ s}$

Using the rotational kinematic equation $\omega_f = \omega_0 + \alpha_{wheel} t$: $0 = 100 \text{ rad/s} + \alpha_{wheel} (2 \text{ s})$ $\alpha_{wheel} = -\frac{100}{2} = -50 \text{ rad/s}^2$

The magnitude of angular deceleration is $50 \text{ rad/s}^2$.

3. Calculate the total torque applied by the brakes on the wheels:

The torque exerted by the brakes on the wheels causes their deceleration.

Given: Moment of inertia of each pair of wheels, $I_{wheel\_pair} = 0.3 \text{ kgm}^2$. Since there are two pairs of wheels (front and rear), the total moment of inertia of all wheels is $I_{total\_wheels} = 2 \times I_{wheel\_pair} = 2 \times 0.3 \text{ kgm}^2 = 0.6 \text{ kgm}^2$.

The torque $\tau_{brakes}$ is given by $\tau = I \alpha$: $\tau_{brakes} = I_{total\_wheels} \times |\alpha_{wheel}| = 0.6 \text{ kgm}^2 \times 50 \text{ rad/s}^2 = 30 \text{ Nm}$

4. Determine the torque exerted by the wheels on the car body:

By Newton's third law, the torque exerted by the brakes on the wheels is equal in magnitude and opposite in direction to the torque exerted by the wheels on the car body. This torque is what causes the car body to rotate about its center of mass.

So, the torque on the car body, $\tau_{car} = 30 \text{ Nm}$.

5. Calculate the angular acceleration of the car body:

Given: Moment of inertia of the car along with the driver about its center of mass, $I_M = 80 \text{ kgm}^2$.

Using Newton's second law for rotation, $\tau = I \alpha$: $\alpha_{car} = \frac{\tau_{car}}{I_M} = \frac{30 \text{ Nm}}{80 \text{ kgm}^2} = \frac{3}{8} = 0.375 \text{ rad/s}^2$

6. Calculate the angle of rotation of the car:

The car body starts rotating from rest (its initial angular velocity about its center of mass is zero).

Given: Initial angular velocity of car body, $\omega_{car,0} = 0 \text{ rad/s}$ Angular acceleration of car body, $\alpha_{car} = 0.375 \text{ rad/s}^2$ Time, $t = 2 \text{ s}$

Using the rotational kinematic equation $\theta = \omega_{car,0} t + \frac{1}{2} \alpha_{car} t^2$: $\theta = (0 \text{ rad/s})(2 \text{ s}) + \frac{1}{2} (0.375 \text{ rad/s}^2) (2 \text{ s})^2$ $\theta = 0 + \frac{1}{2} \times 0.375 \times 4$ $\theta = 0.375 \times 2 = 0.75 \text{ radians}$

The car will rotate by an angle of $0.75$ radians about its centre of mass in the 2 second interval.