Question

A thin stick of mass 0.5 kg and length

L = 0.8 m

is attached to the rim of a metal disk of mass

M = 5.0 kg

and radius

R = 0.4 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.)

Answer 1

The initial potential energy is given by, E = mg (5) + Me (L+R) =f(0.5 kB)(9.81m's») 0.9")) +(5.0 kg)(9.81 m/s?)(0.8m +0.4 m)) =1.962 J + 58.86 =60.8 J The final kinetic energy at the bottom is given by, E = (1 tex +143)2 But, = (0,5 kg)(0.8m) = 0.16 kg-m? lazs == MR? +M(R+L)? = (5kg)(0.4 m)? +(5kg)(0.4 m +0.8m)? = 7.6 kg-m? Thus, from energy conservation, E, = E 3(1.+123 )@?= 60.85 (0.16 kg-m?+ 7.6 kg-m?) oʻ=60.85 0 = 3.96 rad's Thus, linear speed is, v=OR+L) = (3.96 rad's)(1.2 m) = 4.75 m/s

Toner = la Here, AR+L) a=2 me{j}+ Mg(L+R) =+(2+2) 7.6 kg-m? =(Istick +I dist) (0.8 m +0.4 m) 7.6 kg-m²=(0.16 kg-m² +7.6 kg -mº)(0.8 m +0.4 m) a=1.17 m/s? Acceleration is given by. 12 a=RI = (4.75 m/s) 1.2 m =18.8 m/s2