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

L=0.27m

m=1.20kg

d= 0.1m

M= 2.00kg

PE= mg(L/2) + Mg(L + d/2)
PE= (1.20)(9.80)(.27/2) + (2.00)(9.80)(.27+ .1/2)
PE= 1.587 J + 6.272 J
PE= 7.859 J
That is also your rotational kinetic energy.

Now we need to find Itotal.
Irod= (1/3)ML^2= (1/3)(1.20)(.27^2)= 0.02916
Iball= (2/5)MR^2+M(L+d/2)^2= (2/5)(2.0)(.05^2)+(2.0)[(.27+(.1/2))^2= 0.002+0.0248=.0268
Itotal= 0.02916+.0268
Itotal= 0.05596

Now that we have theItotal of the system we can plug it up into the rotational kinetic energy.
K=(1/2)(Itotal)(w^2)
7.859 J = (1/2) (0.05596) (w^2)
7.859 J = 0.02798 (w^2)
Square root of (280.879) = w (angular speed of the ball and rod)
w= 16.75 rad/s

v=w(L+d/2)
v= (16.75rad/s)(.32 m)
v=5.36 m/s

If the ball has fallen freely v = sqrt(2gh) = 2.504m/s

So vfall/vswing = 2.505/5.36 = 0.467

Answers:

(A)7.859 J

(B)16.75 rad/s

(C)5.36 m/s

(D)0.467

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