An green hoop with mass mh = 2.5 kg and radius Rh = 0.16 m hangs from a string t

Question

An green hoop with mass mh = 2.5 kg and radius Rh = 0.16 m hangs from a string that goes over a blue solid disk pulley with mass md = 2 kg and radius Rd = 0.1 m. The other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass ms 3.6 kg and radius Rs 0.24 m. The system is released from rest. 1) What is magnitude of the linear acceleration of the hoop? m/s Submit You currently have O submissions for this question. Only 10 submission are allowed You can make 10 more submissions for this question. 2) What is magnitude of the linear acceleration of the sphere? m/s2 Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question.

Explanation / Answer

F net = m*g = 2.5*9.81 = 24.525 N

I = 7/5*mR^2
M = 7/5*m

a = F/m = 24.525/(2.5+ 1/2*2 + 7/5*3.6)

= 2.872 m/s^2

1)

linear acceleration of hoop= 2.872 m/s^2

2)

linear acceleration of sphere = 2.872 m/s^2

3)
Angular acceleration = a/R = 2.872 /0.1 = 28.72rad/s^2

4)

angular acceleration sphere = a/R = 2.872/0.24 = 11.97rad/s^2

5)

tension between pulley and sphere = M*a = 7/5*3.6*2.872 = 14.475 N

6)

tension between hoop and pulley = m(hoop) (g - a)
= 2.5(9.81 - 2.872)

= 17.32 N

7)

t = sqrt(2s/a) = sqrt(2*1.59/2.872)
t = 1.0522 seconds

8)

v = sqrt(2as) = sqrt(2*2.872*1.59) = 3.022 m/s

9)

omega = v/R = 3.022 /0.24 = 12.592 rad/s

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