A string is wrapped tightly around the circumference of a wheel with a mass of 1

Question

A string is wrapped tightly around the circumference of a wheel with a mass of 1.23 kg and a radius of .324 m. The string is attached to a block with a mass of 2.48 kg which is .34 m above the floor. When the mass is released the wheel rotates freely as the string unwinds without slipping. a) What is the initial energy of the system? b) What is the moment of inertia of the wheel? c) Write the expression for the total energy of the system just before the block hit the floor. d) What is the velocity of the block just before it hits the floor? e) What is the angular velocity of the wheel just before the block hits the floor?

Explanation / Answer

a)

Initial energy = mgh = 2.48*9.81*0.34 = 8.272 J

b)

MOI of wheel I = 1/2*MR^2 = 1/2*1.23*0.324^2 = 0.0646 kg-m^2

c)

Total energy = Rotational KE of wheel + Translational KE of block = 1/2*I^2 + 1/2*mv^2

d)

Torque = T*R = I

where T is tension in string.

Also, ma = mg - T

So, T = m(g-a)

m(g-a) *R = I

Also, = a/R

So, m(g-a) *R = I*a/R

mg = (m + I/R^2)a

a = mg / (m + I/R^2)

a = 2.48*9.81 / (2.48 + 0.0646/0.324^2)

a = 7.86 m/s^2

Using, v^2 = u^2 + 2ah

v^2 = 0 + 2*7.86*0.34

v = 2.32 m/s

e)

= v/R = 2.32/0.324 = 7.136 rad/s

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