Question

A wheel starts from rest under the influence of a
constant torque and turns through 500 rad in 10s.
(a) What is its angular acceleration? (b) What is
its angular velocity after 10s?

Answer 1

Angular Velocity is 80 rad/s

Explanation:

The initial angular velocity $\omega = 0$ and Angular velocity after 10 s,

$\omega = \frac {400}{10} = 40\ rads^-^1$

Therefore angular acceleration, $\alpha = \frac {\omega - \omega_0}{t}$

= $\frac {40 - 0}{10} = 4\ rads^-^2$

Now angular velocity after 20 s is given by,

$\omega = \omega_0 + \alpha t$

$\omega = 0 + 4 \times 20 = 80 rads^-^1$

Answer 2

The angular acceleration is $5 rad/s^2$.

The angular velocity after 10 s is $50 rad/s^2$.

Explanation:

Given initial angular velocity, $\omega_{0} =0$.

Wheel turns through 500 rad in 10 s.

So final angular velocity after 10 s,

$\omega_{f}=\frac{ \theta}{t} =\frac{500}{10} =50 rad/s$

(a )To calculate the angular acceleration use equation of rotational motion,

$\omega_{f} =\omega_{0} +\alpha t$.

Here, \alpha is angular acceleration.

Substitute the values, we get

$50 =0+\alpha \times 10s$

$\alpha=\frac{50}{10} =5 rad/s^2$

Thus, the angular acceleration is $5 rad/s^2$

(b) To calculate the angular velocity after 10 s again use equation of rotational motion,

$\omega_{f} =\omega_{0} +\alpha t$.

Substitute the values, we get

$\omega_{f} =0+5rad/s^2 \times10s$

$\omega_{f} =50 rad/s^2$.

Thus, the angular velocity after 10 s is $50 rad/s^2$.

Topic : Equation of rotational motion.

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