An green hoop with mass mh = 2.8 kg and radius Rh = 0.17 m hangs from a string t

Question

An green hoop with mass mh = 2.8 kg and radius Rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass md = 2.4 kg and radius Rd = 0.08 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 = 3.3 kg and radius Rs = 0.18 m. The system is released from rest.

What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.54 m)?

Explanation / Answer

the only force on the system is the weight of the hoop
F net = 2.8kg*9.81m/s^2 = 27.468 N

The inert masses that must be accelerated are the ring, pulley, and the rolling sphere.

The mass equivalent of M the pulley is found by
torque ? = F*R = I*? = I*a/R
F = M*a = I*a/R^2 -->
M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m

The mass equivalent of the rolling sphere is found by:
the sphere rotates around the contact point with the table.
So using the theorem of parallel axes, the moment of inertia of the sphere
is I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2
M = 7/5*m

the acceleration is then
a = F/m = 27.468/(2.8 + 1/2*2.4 + 7/5*3.3) = 27.468/8.62 = 3.18 m/s^2
----------------
linear acceleration of hoop= 3.18 m/s^2

lin. accel. of sphere = 3.18 m/s

angular acceleration disk pulley:
? = a/R = 3.18 /0.08 = 39.75 rad/s^2

ang. acceleration sphere = a/R = 3.18 /0.18 = 17.66 rad/s^2

tension between pulley and sphere = M*a = 7/5*3.3*3.18 = 14.691 N

tension between hoop and pulley = m(hoop) (g - a)
= 2.8(9.81 - 3.18) = 18.56 N

s = 1/2*at^2 --> t = ?(2s/a) = ?(2*1.54/3.18)
--> t = 0.9841 seconds

v = ?(2as) = ?(2*3.18*1.54) = 3.129 m/s

? = v/R = 3.129/0.18 = 17.38rad/s

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