You are sitting on the edge of a merry-go-round of radius 2.50 m. When you are a

Question

You are sitting on the edge of a merry-go-round of radius 2.50 m. When you are at the eastern-most point of your motion, you have angular velocity of 42 rpm and you start to drag your foot on the ground. By the time you have made half a revolution, you are moving with angular velocity of 21 rpm. Assuming constant angular acceleration, what are the magnitudes and direction of your linear acceleration when it is

(a) at the northern-most point of its trajectory after making a quarter of a revolution, and

(b) at the western-most point of its trajectory after making half of a revolution?

(c) When and where does the merry-go-round stop?

Explanation / Answer

Initial angular speed, wi = 42 rpm = 42 x 2pi rad/60 sec = 4.40 rad/s

final angular speed, wf = 21 rpm = 2.20 rad/s

and during this deceleration, pi rad angle is revolved.

applying, wf^2 - wi^2 = 2 (alpha) (theta)

2.20^2 - 4.40^2 = 2(alpha) (pi)

alpha = - 2.31 rad/s^2

(A) angular acceleration (alpha) is constant.

hence tangential acc., a_t = alpha *r = 2.31 * 2.50 = 5.78 m/s^2

direction of a_t (tangential and opposite of direction of velocity) -> east

a_t = 5.78i m/s^2

radial acc. = w^2 r   and direction will be towards the centre.

at this moment, angle turrned is pi/2 rad.

applying wf^2 - wi^2 = 2 alpha theta   to find angular speed at this moment,

w^2 - 4.40^2 = (2)(-2.31)(pi/2)

w = 3.48 rad/s

a_r = 3.48^2 (2.50) = 30.26 m/s^2 (-j) = - 30.26 j m/s^2

net linear acc. at this point, a = a_t + a_r = 5.78i - 30.26j m/s^2 ......This is the vector form of linear acc.

magnitude = sqrt(5.78^2 + 30.26^2) = 30.81 m/s^2

direction = tan^-1(30.26 / 5.78) = 79.2 deg south of east.

(B) magnitude of a_t will be same throughout the motion.

for western most point,

a_t = 5.78j

a_r = w^2 r (i)

at this point, w = 21 rpm = 2.20 rad/s

a_r = 2.20^2 (2.50) i = 12.1 i m/s^2

a = a_t + a_r   = 12.1i + 5.78j   m/s^2   { vector form of linear acc. }

magnitude = sqrt(12.1^2 + 5.78^2) = 13.41 m/s^2

direction = tan^-1(5.78/12.1) = 25.5 deg north of east.

(c) merry-go-round will stop when its angular speed will become zero.

applying, wf^2 - wi^2 = 2 (alpha) (theta)

as it stops, wf = 0

0^2 - 4.40^2 = 2(-2.31) (theta)

theta = 4.19 rad     (so it will stop after it has turned 4.19 rad)

IN revolutions. rev = 4.19/ 2pi = 1.5 revolutions ,.....Ans

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