Question

Torque of 50 n m wheel at rest rotates through 200 rad 10 sec angular acc is

Answer 1

here torque = 50Nm

ω initial = 0

∅ = 200rad.

t= 10sec

by law of rotation

∅= ωinitial t + 1/2 α t^2

200 = 1/2 α× 10×10

α= 4 rad/sec^2

Answer 2

The angular acceleration of the wheel is $4\ rad/s^2$. H

Explanation:

Given that,

Torque acting on the wheel, $\tau=50\ N-m$

Angular displacement, $\theta=200\ rad$

Time, t = 10 s

Initial velocity of the wheel, $\omega_i=0$

We need to find the angular acceleration of the wheel. It can be calculated using second equation of motion as :

$\theta=\omega_i+\dfrac{1}{2}\alpha t^2$

$\theta=\dfrac{1}{2}\alpha t^2$

$\alpha =\dfrac{2\theta}{t^2}$

$\alpha =\dfrac{2\times 200}{(10)^2}$

$\alpha =4\ rad/s^2$

So, the angular acceleration of the wheel is $4\ rad/s^2$. Hence, this is the required solution.

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Rotational kinematics

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