Question

On the application of constant torque a wheel is turned from rest through 400radian in 10second find angular acceleration. If same torque continue to act what will be the angular velocity of wheel after 20 second from start

Answer 1

Answer:

Angular acceleration = $8\ rad/s^2$

Angular velocity after 20 s = 160 rad/s

Explanation:

Case I:

• $\theta$ = angular turn of the wheel = 400 rad
• t = time interval = 10 s
• $\omega_i$ = initial angular velocity of the wheel = 0 rad/s

Since the wheel is acted upon by a constant torque, it has a constant angular acceleration. So, we have

$\theta = \omega_i t +\dfrac{1}{2}\alpha t^2\\\Rightarrow 400=0\times 10+\dfrac{1}{2}\alpha 10^2\\\Rightarrow 400=50\alpha\\\Rightarrow \alpha = \dfrac{400}{50}\\\Rightarrow \alpha = 8\ rad/s^2$

Hence, the angular acceleration of the wheel is $8\ rad/s^2$.

Case II:

time interval = t = 20 s

The final angular acceleration of the wheel is calculated using the following relation:

$\omega_f = \omega_i +\alpha t\\\Rightarrow \omega_f = 0 +8\times 20\\\Rightarrow \omega_f =160\ rad/s$

Hence, the angular velocity of the wheel is 160 rad/s.

Answer 2

Answer:

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