Question

A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so small that it can be ignored. A sandbag of mass m is dropped onto the merry-go-round, at a position designated by r. The sandbag does not slip or roll upon contact with the merry-go-round

Rank the following different combinations of and on the basis of the angular speed of the merry-go-round after the sandbag "sticks" to the merry-go-round.

m1=10kg m2=10kg m3=20kg m4=15kg m5=10kg m6=40kg
r1=.50R r2=.25R r3=.25R r4=.75R r5=1.0R r6=.25R

Rank from largest to smallest. To rank items as equivalent, overlap them.

Answer 1

From law of conservation of angular momentum,

$$ I_{1} \omega_{1}=I_{2} \omega_{2} $$

\(I_{2} \propto \frac{1}{\omega_{2}}\)

And,

$$ I_{2}=I_{1}+m r^{2} $$

\(I_{2} \propto m r^{2}\)

Now,

(a) For \(m=10 \mathrm{~kg}, r=0.50 R\)

\(m r^{2}=\frac{10}{4}=\frac{40}{16}\)

(b) For \(m=20 \mathrm{~kg}, r=0.25 R\)

\(m r^{2}=\frac{20}{16}\)

(c) For \(m=10 \mathrm{~kg}, r=1.0 R\)

\(m r^{2}=\frac{160}{16}\)

(d) For \(m=15 \mathrm{~kg}, r=0.75 R\)

\(m r^{2}=\frac{135}{16}\)

(e) For \(m=10 \mathrm{~kg}, r=0.25 R\)

$$ m r^{2}=\frac{10}{16} $$

(f) For \(m=40 \mathrm{~kg}, r=0.25 R\)

$$ m r^{2}=\frac{40}{16} $$

More \(m r^{2}\) means more \(I_{2}\), and more \(I_{2}\) means less \(\omega_{2}\). The order is, \((e)>(b)>(a)=(f)>(d)>(c)\)

Answer 2

1, 2, 4, 5, 3, 6

 

Answer 3

From law of conservation of angular momentum

The order is, (e)>(b)>(a)=(f)>(d)>(c)