c) 0,035 Extra Credit. (5 points) A green hoop with mass mh- 2.8 kg hangs from a

Question

c) 0,035 Extra Credit. (5 points) A green hoop with mass mh- 2.8 kg hangs from a string that goes over a blue solid disk pulley with mass md 1.9 kg and radius Rd-0.09 m. The other end of the string is attached to a massless axel through the center of an orange sphere on a flat frictionless horizontal surface and has mass ms 3.8 kg and radius R-0.23 m. The system is released from rest. What is the acceleration of the green hoop if the sphere is pulled along the surface so that the sphere no longer rotates? 2.8k

Explanation / Answer

The only force on the system is the weight of the hoop,

F net = mh*g = 2.8 kg*9.81 m/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, T = F*R = I*a = I*a/R

F = M*a = I*a/R^2

M = I/R^2 = 21/2*md*R^2/R^2 = 1/2*md

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*msR^2 for rotation about the center of mass + msR^2 for the distance of the axis of rotation from the center of mass of the sphere.

I = 7/5*msR^2

M = 7/5*ms

the acceleration is then

a = F/m

= 27.468/(2.8 + 1/2*1.9 + 7/5*3.8)

= 3.028 m/s^2 answer

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