Angreen h 00 with mass m, . 2.8 kg and radis R-0.12 m hags froma string that goe

Question

Angreen h 00 with mass m, . 2.8 kg and radis R-0.12 m hags froma string that goes oer a blee toldea pulley with mass me 2 kg and radlus Rd 0.09 m. The other end of the string is attached to a massdless axel through the center of an orange sphere on a flat hortzontal surface that rolls without slipping and has mass my-4 kg and radius R 0.23 m. The system is released from rest. 1) What is magnitude of the linear acceleration of the hoop? 2) What is magniltude of the linear acceveration of the sphere )what is the magnitude of the angular acceleration of the disk pulley 4) What is the magnitude of the anguiar acceleration of the sphere 5) What is the tensian in the string between the sphere and disk puliey? Q search or enter website name 2 4 5 8

Explanation / Answer

Given,

mh = 2.8 kg ; rh = 0.12 m ; ms = 4 kg ; rs = 0.23 m ; md = 2 kg ; rd = 0.09 m

1)The net force acting on the system is weight of the hoop

F = mh g = 2.8 x 9.8 = 27.44 N

Is = 2/5 m r^ + m r^2 = 7/5 m r^2

M = 7/5 m

a = F/m = 27.44/(2.8 + 1/2 x 2 + 7/5 x 4) = 2.92 m/s^2

Hence, a = 2.92 m/s^2

b)a = 2.92 m/s^2

3)alpha-d = a/rd = 2.92/0.09 = 32.44 rad/s^2

Hence, alpha-d = 32.44 rad/s^2

4)alpha-s = a/rs = 2.92/0.23 = 12.69 rad/s^2

Hence, alpha-s = 12.69 rad/s^2

5)T = Ma

T = 7/5 x 4 x 2.92 = 16.35 N

Hence, T = 16.35 N

6)T' = mh(g - a)

T' = 2.8 (9.81 - 2.92) = 19.29 N

Hence, T' = 19.29 N

7)d = 1.49 m

t = sqrt (2d/a)

t = sqrt(2 x 1.49/2.92) = 1.01 s

Hence, t = 1.01 s

8)v = sqrt (2 a d)

v = sqrt (2 x 2.92 x 1.49) = 2.95 m/s

Hence, v = 2.95 m/s

9)v = rw => w = v/r

w = 2.95/0.23 = 12.83 rad/s

Hence, w = 12.83 rad/s

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