A thin stick of mass 0.3 kg and length L = 0.7 m is attached to the rim of a met

Question

A thin stick of mass 0.3 kg and length L = 0.7 m is attached to the rim of a metal disk of mass M = 4.0 kg and radius R = 0.6 m. The stick is free to rotate around a horizontal axis through its other end (see the following figure).

(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? ___m/s

(b) What is the acceleration (in m/s2) of the center of the disk at the instant the stick is released? (Enter the magnitude.) ___ 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.) ___ m/s2

Answers is

A. 4.83m/s

B. 8.98 m/s2

C.18 m/s2

WHY??

Explanation / Answer

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

Ei = (0.3*9.81*0.35) + (4*9.81*1.3) = 52.04 J

final kinetic energy at the bottom Ef = (1/2)*(Istick + Idisk)*w^2

Istick = (1/3)*m*L^2 = (1/2)*0.3*0.7^2 = 0.0735 kg m^2

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

Idisk = (1/2)*4*0.6^2 + (4*1.3^2) = 7.48 kg m^2

from energy conservation Ef = Ei

(1/2)*(0.0735+7.48)*w^2 = 52.04

angular speed w = 3.712 rad/s

linear speed v = w*(R+L) = 3.712*1.3 = 4.83 m/s

=======================

part B

net torque = I*alpha

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

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

(0.3*9.83*0.35) + (4*9.83*1.3) = (0.0735+7.48)*a/1.3

a = 8.98 m/s^2

=========================

part (C)

acceleration a = v^2/(R+L) = 4.83^2/1.3 = 18 m/s^2

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