Can I get help for this question please.. 96. A thin stick of mass 0.2 kg and le

Question

Can I get help for this question please..
96. A thin stick of mass 0.2 kg and length L = 0.5 m is attached to the rim of a metal disk of mass M = 2.0 kg and radius R = 0.3 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 of the center of the disk whern the stick is vertical? (b) What is the acceleration of the center of the disk at the instant the stick is released? (c) At the instant the stick passes through the vertical? to ab This OpenStax book is available for free at ht ttp.//enx.org/content/col1.2031/1.5

Explanation / Answer

A) By energy conservation,

Decrease in PE = increase in KE

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

0.2*9.8*0.5/2 + 2*9.8*0.8 = 0.5*(1/3*0.2*0.5^2 + 0.5*2*0.3^2+ 2*0.8^2) w^2

16.17 = 0.5*1.38666 w^2

w = sqrt(16.17/0.69333)

= 4.83 rad/s

speed of center of disc= w (L+r) = 4.83*0.8 = 3.864 m/s

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

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

= (0.2*9.8*0.5/2+2*9.8*0.8)/1.38666 *0.8

= 9.33 m/s^2

C) a= 0 because torque is zero

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