Question

A disc spinning at the rate 27.5 rad s is slowed at the rate 10 rad s 2. The time after
which it will come to rest is
JA) 2.75 s (B) 5.5 s (C) 1.25 s (D) 3.5 s
(E) 6.2 s
A)​

Answer 1

Given:

A disc spinning at a rate of 27.5 rad/s. It is slowed down at a rate of 10 rad/s².

To find:

Time after which the disc comes to rest.

Calculation:

Since the angular acceleration is constant, we can easily apply the equations of Rotational Kinematics to solve this type of questions:

Let initial angular velocity be $\omega1$ and angular acceleration be $\alpha$.

$\therefore \: \omega2 = \omega1 + \alpha t$

$= > \: \alpha t = \omega2 - \omega1$

$= > \: t = \dfrac{ \omega2 - \omega1}{ \alpha }$

$= > \: t = \dfrac{ 0- 27.5}{ - 10 }$

$= > \: t = \dfrac{ - 27.5}{ - 10 }$

$= > \: t = \dfrac{ 27.5}{ 10 }$

$= > \: t = 2.75 \: sec$

So, final answer is:

$\boxed{ \sf{ \blue{ \large{\: t = 2.75 \: sec}}}}$