A thin, cylindrical rod = 27.0 cm long with a mass m = 1.20 kg has a ball of dia

Question

A thin, cylindrical rod = 27.0 cm long with a mass m = 1.20 kg has a ball of diameter d = 10.00 cm and mass M = 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top as shown in the figure below. The combination is free to pivot about the bottom end of the rod after being given a slight nudge. (a) After the combination rotates through 90 degrees, what is its rotational kinetic energy? (b) What is the angular speed of the rod and ball? (c) What is the linear speed of the center of mass of the ball? (d) How does it compare with the speed had the ball fallen freely through the same distance of 32.0 cm? vswing is vfall by

Explanation / Answer

Centre of mass:

d = (m*d) / m

= [1.20*27/2 + 2(27+10/2)] / (1.20 + 2)

= 1.20(13.5)+(2*32) / 3.20

= (16.2+64)/3.20=80.2/3.20=25.06 cm

d=25.06 cm =0.2506 m

Ek = Ep Ek = m*g*h

= (1.20+2)*9.8*0.2506

Ek= 7.907 J

(B) from the equation Ek = 0.5*I*2

7.907 = 0.5*m*r2*2

7.907 = 0.5*(1.20+2)*0.2506*2

2=(7.907)/(0.5*3.20*0.25062)

2=7.907/0.10048=78.62 r/s

= 8.87 r/s

(C) Linear Speed Formula in terms of angular speed is given by V=wr

here, is angular speed and
            r is the radius of circular path.
The Linear speed formula is used to calculate the linear speed of any given body if its angular velocity and radius of circular path are given. Linear speed is expressed in meter per speed (m/s).

Now Linear Speed V=wr=8.87*(0.2506/2)=1.11 m/s

(D) the speed had the ball fallen freely through the same distance of 32.0 cm

mgh = 1/2 I 2
(1.02+2)9.8*0.2506=(0.5)(0.32)(8.87)2

(3.02)*9.8*0.2506=(0.5*0.32*78.67)

7.41~= 12.5(here 12.5>7.41)

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