Question

A thin stick of mass 0.4 kg and length L = 0.3 m is attached to the rim of a metal disk of mass M = 4.0 kg and radius R = 0.2 m. The stick is free to rotate around a horizontal axis through its other end (see the following figure). m VE? (a) If the combination is released with the stick horizontal, what is the speed (in m/s) of the center of the disk when the stick is vertical? X m/s (b) What is the acceleration (in m/s) of the center of the disk at the instant the stick is released? (Enter the magnitude.) x m/s2 (c) What is the acceleration in m/s2) of the center of the disk at the instant the stick passes through the vertical? (Enter the magnitude.) x m/s2

Can you please help me understand this question, I have a test soon and I don't quite know how to use the energy equation.

Answer 1

(a) Intial potential energy

Ei = m*g*(L/2) + M*g*(L+R)

Ei = (0.4*9.8*0.15) + (4*9.8*0.5) = 20.188 J

Final kinetic energy at the bottom

Ef = (1/2)*(Istick + Idisk)*w^2

Istick = (1/2)*m*L^2 = (1/2)*0.4*0.3^2 = 0.018 kg m^2

Idisk = (1/2)*M*R^2 + M*(R+L)^2

Idisk = (1/2)*4*0.2^2 + (4*0.5^2) = 1.08 kg m^2

from energy conservation Ef = Ei

(1/2)*(0.018 + 1.08)*w^2 = 20.188

angular speed w = 6.06 rad/s

linear speed v = w*(R+L) = 6.06 *0.5 = 3.03 m/s

(b) net torque = I*alpha

alpha = angular acceleration = a/(R+L)

m*g*L/2 + M*g*(R+L) = I*a/(R+L)

(0.4*9.81*0.15) + (4*9.81*0.5) = (0.018+1.08)*a/0.5

a = 9.19 m/s^2

(c) acceleration a = v^2/(R+L) = 3.03^2/0.5 = 18.36 m/s^2

Kindly upvote :)