Question

The speed of a motor decreases uniformly with time from 1200 rpm to 600 rpm i. 20 seconds. The total no.Of rotations it makes before coming to rest is

Answer 1

Answer:

$n = 98 \pi rads$

Explanation:

Given

$v_1 = 1200 rpm = 20 \pi rads \: {s}^{-1}$

$v_2 = 600 rpm = 10 \pi rads \:{s}^{-1}$

$s = 2 \pi rads$

$t = 20 sec$

To Find

Total no. of rotations

Solution

v = u + at

20 = 10 + (-a) 20

20 = 10 -20a

20a = -10

a = -0.5 π rads/s²

Now,

$\boxed{ Sn = ut+\dfrac{1}{2}a{t}^{2} }$

$2 \pi n = 10 \pi rads \: {s}^{-1}(20s)+\dfrac{1}{2}(-0.5 π rads\: {s}^{-2}) {20s}^{2}$

$2 \pi n= 200 \pi rads - 100 \pi rads$

$n= 100 \pi rads - 2\pi rads$

$n = 98 \pi rads$