An green hoop with mass m h = 2.4 kg and radius R h = 0.14 m hangs from a string

Question

An green hoop with mass mh = 2.4 kg and radius Rh = 0.14 m hangs from a string that goes over a blue solid disk pulley with mass md = 2.3 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 = 3.7 kg and radius Rs= 0.23 m. The system is released from rest.

1) What is magnitude of the linear acceleration of the hoop?

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

Can anyone explain how this is work?

Explanation / Answer

The only force on the system is the weight of the hoop
F net = 2.4kg*9.81m/s^2 = 23.544 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 = 1/2*2.3

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 = 7/5*3.7

the acceleration is then
a = F/m = 23.544/(2.4 + 1/2*2.3 + 7/5*3.7) = 27.468/8.73 = 2.696 m/s^2
----------------
therefore, linear acceleration of hoop= 2.696 m/s^2

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