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Question: On the application of a constant torque, a wheel is turned from rest through \(400\,radians\) in \(1...

On the application of a constant torque, a wheel is turned from rest through 400radians400\,radians in 10s10\,s.
(i) Find angular acceleration.
(ii) If the same torque continues to act, what will be the angular velocity of the wheel after 20s20\,s from start?

(A) (i) 8rads28\,rad{s^{ - 2}}, (ii) 160rads1160\,rad{s^{ - 1}}
(B) (i) 16rads216\,rad{s^{ - 2}}, (ii) 160rads1160\,rad{s^{ - 1}}
(C) (i) 8rads28\,rad{s^{ - 2}}, (ii) 80rads180\,rad{s^{ - 1}}
(D) (i) 4rads24\,rad{s^{ - 2}}, (ii) 80rads180\,rad{s^{ - 1}}

Explanation

Solution

In order to calculate the angular acceleration use the angular displacement formula below and substitute the known values. From the angular acceleration obtained, use the angular velocity formula to know the value of the angular velocity.
Useful formula:
(1) The angular displacement is given by,
θ=ω0t+12αt2\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}
Where θ\theta is the angular displacement, ω0{\omega _0} is the initial angular velocity, α\alpha is the angular acceleration and tt is the time taken.
(2) The angular velocity is given by
ω=αt\omega = \alpha t
Where ω\omega is the angular velocity.

Complete step by step solution:
It is given that the
Wheel is turned through the distance in angular displacement, θ=400rad\theta = 400\,rad
Time taken for the turning, t=10st = 10\,s
The angular acceleration is obtained by using the formula (1),
θ=ω0t+12αt2\theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}
The angular velocity at the initial is considered as 00.
400=(0)(10)+12α(10)2400 = \left( 0 \right)\left( {10} \right) + \dfrac{1}{2}\alpha {\left( {10} \right)^2}
By simplifying the above equation, we get
400=12(100α)400 = \dfrac{1}{2}\left( {100\alpha } \right)
α=800100\alpha = \dfrac{{800}}{{100}}
By performing the division in the above step,
α=8rads2\alpha = 8\,rad{s^{ - 2}}
Hence the angular acceleration obtained is 8rads28\,rad{s^{ - 2}}
(ii) Using the formula(2) for calculating the angular velocity at 20s20\,s time.
ω=αt\omega = \alpha t
Substituting the values,
ω=8×20\omega = 8 \times 20
ω=160rads1\omega = 160\,rad{s^{ - 1}}
Hence the angular velocity at 20s20\,s is 160rads1160\,rad{s^{ - 1}}.

Thus the option (A) is correct.

Note: The formula (1) of the angular displacement is applicable only at the time of the constant torque. In case if the torque is non constant and if it varies, then the angular acceleration also varies with the time. This is because the torque is the ability of the body to cause angular rotation.