Question

a rotating wheel requires 3.00 s to rotate through 37.0 revolutions. its angular speed at the end of the 3.00 s interval is 98.1 rad/s. what is the constant angular acceleration of the wheel?

Answer 1

13.74

Step-by-step explanation:

Above equation can be solved using the equation of rotational motion with constant angular acceleration

`\theta = \omega_0 t + (1)/(2) \alpha t^2`

`\omega = \omega_0 + \alpha t`

As per the question,

`\theta = 74\pi (converting to radians) \\\\t = 3 \\\\\omega_((t = 3)) = 98.1`

Substituting above values we get the equations,

`74\pi = \omega_0 3 + (1)/(2)\alpha(3)^2\\` … (1)

`98.1 = \omega_0 + \alpha 3` … (2)

Solve these equations simultaneously, to get the value of alpha substitute value of
`\omega_0` from equation 2, in the equation 1

`74\pi = (98.1 - 3\alpha)3 + (1)/(2)\alpha 9\\(9)/(2)\alpha = 98.1*3 - 74\pi\\`

On solving, we get

`\alpha = 13.74`

Hopefully, This answer helped you!!