Question

A coin is placed on the horizontal surface of a rotating disc. The distance of the coin from the axis is 1 m and coefficient of friction is 0.5. if the disc starts from rest and is given an angular acceleration 1√2rad/sec212rad/sec2, the number of revolutions through which the disc turns before the coin slips is​

Answer 1

Answer:

Coin will start slipping after disc will complete 0.56 turns

Explanation:

For slipping of coin we can say that

force of friction is counterbalanced by pseudo force on it

so here we will have

$m\omega^2 r = \mu m g$

so we will have

$\omega = \sqrt{\frac{\mu g}{r}}$

$\omega = \sqrt{\frac{0.5 \times 10}{1}}$

$\omega = 2.24 rad/s$

now we know that angular acceleration is given to us

so we have

$\omega^2 - \omega_i^2 = 4\pi \alpha N$

$2.24^2 - 0 = 4\pi(\frac{1}{\sqrt2})N$

$N = 0.56 turns$

#Learn

Topic : circular motion

https://brainly.in/question/6961555

Answer 2

Answer:

7/4π is the answer