A 2.00 kg stone is tied to a thin, light wire wrapped around the outer edge of t

Question

A 2.00 kg stone is tied to a thin, light wire wrapped around the outer edge of the uniform 9.00 kg cylindrical pulley shown in the figure below (Figure 1) . The inner diameter of the pulley is 55.0 cm , while the outer diameter is 1.00 m . The system is released from rest, and there is no friction at the axle of the pulley.

A) Find the acceleration of the stone. (Express your answer in meters per second squared to three significant figures.)

B) Find the tension in the wire. (Express your answer in newtons to three significant figures.)

C) Find the angular acceleration of the pulley. (Express your answer in radians per second squared to three significant figures.)

Explanation / Answer

Torque on the pulley= R*T = I* = I*a/R

R is the outer radius of the pulley
T is the tension off the wire
I = moment of inertia of the pulley
= angular acceleration
a = tangential acceleration

By vertical force balance
Tension T=m*( g - a).
this makes the equation to
(mg - ma)*R = I*a/R ( m = mass of the stone)
mg - ma = Ia/R2 --> solve for a
mg = a(m + I/R2)
a = mg/ (m + I/R2)

I of the pulley is
I = m*(r12 + r22)/2)
I = 2*(0.52 + 0.2752)/2)
I = 0.326 kgm2

a = 2*9.81/(2 + 0.326/0.5^2)
a = 5.938 m/s^2 <--- acceleration of the stone

= a/R = 5.938/0.5 = 11.876 rad/s2 <--- angular acceleration

T = mg - ma = 2(9.81 - 5.938) = 7.744 N <--- tension

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