A thin stick of mass 0.4 kg and length L = 0.6 m is attached to the rim of a met

Question

A thin stick of mass 0.4 kg and length L = 0.6 m is attached to the rim of a metal disk of mass M = 5.0 kg and radius R-0.5 m. The stick is free to rotate around a horizontal axis through its other end (see the following figure). v=? (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 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.) (c) m/s2

Explanation / Answer

A) By energy conservation,

Decrease in PE = increase in KE

mgL/2 + Mg(L+R) = 0.5 iw^2

0.4*9.8*0.6/2 + 5*9.8*1.1 = 0.5*(1/3*0.4*0.6^2 + 0.5*5*0.5^2+ 5*1.1^2) w^2

55.076= 0.5*6.723 w^2

w = sqrt(55.076/(0.5*6.723))

= 4.048 rad/s

speed of center of disc= w (L+r) = 4.048*1.1= 4.45 m/s

B) acceleration =alpha*(L+r) = torque/i *(L+r)

= (mgL/2 + Mg(r+L))/i *(L+r)

= (0.4*9.8*0.6/2+5*9.8*1.1)/6.723 *1.1

= 9.01 m/s^2

C) a= 0 because torque is zero

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