Question

A merry-go-round initially at rest at an amusement park begins to rotate at time t=0. The angle through which it rotates is described by θ(t)=πk(t+ke−t/k), where k is a positive constant, t is in seconds, and θ is in radians. The angular velocity of the merry-go-round at t=T is

Answer 1

Answer:

Angular velocity of merry-go-round is πk - 1 at t= T

Explanation:

From the question it is given that

$\theta(t) = \pi k(t+k_e-\frac{t}{k} )$ ..........................(1)

since mathematically, angular velocity is defined as

$\omega(t) = \frac{d\theta(t)}{dt}$ ........................(2)

on substituing the value of θ(t) from equation 1 in equation (2) we get

$\omega(t) = \frac{d\theta(t)}{dt}$ = $\frac{d\pi k (t + k_e - \frac{t}{k} )}{dt}$   ............................(3)

on differentiating equation (3) with respect to time we get

ω(t) = πk(1 -$\frac{1}{k}$) = πk - 1 angular velocity of merry-go-round

Therefore, angular velocity of merry-go-round is πk - 1 at t= T