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Solved: A child places a picnic basket on the outer rim of

Chapter 6, Problem 96

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QUESTION:

A child places a picnic basket on the outer rim of a merry-goround that has a radius of 4.6 m and revolves once every 30 s.

(a) What is the speed of a point on that rim?

(b) What is the lowest value of the coefficient of static friction between basket and merrygo-round that allows the basket to stay on the ride?

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QUESTION:

A child places a picnic basket on the outer rim of a merry-goround that has a radius of 4.6 m and revolves once every 30 s.

(a) What is the speed of a point on that rim?

(b) What is the lowest value of the coefficient of static friction between basket and merrygo-round that allows the basket to stay on the ride?

ANSWER:

Step 1 of 3

You need two concepts to solve this problem.

One is that of circular motion and another of static friction.

Circular motion:

The speed of the object at any point in circular motion is given by:

\(v=\frac{\text { circumference }}{\text { period of revolution }}\)

\(\text { Circumference }=2 \pi R\)

\(\text { period of revolution }=t\)

\(v=\frac{2 \pi R}{t} \ldots \ldots \ldots \ldots \ldots \ldots(1)\)

Static Friction:

For the basket not to slip, the minimum static  frictional force should be equal to the centripetal force acting of the basket.

\(\begin{array}{l}
f_{\text {smin }}=F_{\text {centripetal }} \\
f_{\text {smin }}=\frac{m v^{2}}{R}
\end{array}\)

We know that frictional force is given by,

\(f_{\text {smin }}=\mu_{s} m g\)

Therefore:

\(\begin{array}{l}
f_{\text {smin }}=\mu_{s} m g=\frac{m v^{2}}{R} \\
\mu_{s} m g=\frac{m v^{2}}{R} \dots \dots (2)
\end{array}\)

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