Question

A thin spool with a mass of 2.0kg and a radius of 15cm is hung in a fixed position so that it is free to rotate. A massless string is wrapped around it with a 0.80kg mass tied at the end. When the mass is released, the spool begins to unwind.

What is the rotational inertia of the spool? (You can treat it as a thin disk.)

What is the kinetic energy and angular momentum of the spool 2.0s after the

hanging mass is released?

What is the kinetic energy and angular momentum of the spool after 3 full

rotations?

Answer 1

Moment of Inertia of spool, I = 0.5*M*R^2

= 0.5*2*0.15^2

= 0.0225 kg.m^2 <<<<<<<<---------------Answer

Let T is the tension in the string and a is the acceleration of te hangung mass.

Net force acting on hanging mass, Fnet = m*g - T

m*a = m*g - T

T = (m*g - m*a) ---(1)

Net Torque acting on spool, Torque = I*alfa

T*R = I*a/R

T = I*a/R^2

T = 0.5*M*R^2*a/R^2

T = 0.5*M*a

m*g - m*a = 0.5*M*a

a = m*g/(m + 0.5*M)

= 0.8*9.8/(0.8 + 0.5*2)

= 4.36 m/s^2

so, alfa = a/R

= 4.36/0.15

= 28 rad/s^2

after 2 s, the angular speed of spool, w = wo + alfa*t

= 0 + 29*2

= 58 rad/s

so, kinetic enrgy, KE = 0.5*I*w^2 = 0.5*0.0225*58^2 = 37.85 J <<<<---Answer

angular momentum, L = I*w = 0.0225*58 = 1.305 kg.m^2/s <<<<---Answer

after competing 3 rotations
apply,

w^2 - wo^2 = a*lafa*theta

w = sqrt(2*alfa*theta)

= sqrt(2*28*3*2*pi)

= 32.5 rad/s

so, kinetic enrgy, KE = 0.5*I*w^2 = 0.5*0.0225*32.5^2 = 11.9 J <<<<---Answer

angular momentum, L = I*w = 0.0225*32.5 = 0.731 kg.m^2/s <<<<---Answer