Chapter 8: Q. 52 (page 201)

Suppose you swing a ball of mass m in a vertical circle on a string of length L. As you probably know from experience, there is a minimum angular velocity ωmin you must maintain if you want the ball to complete the full circle without the string going slack at the top.
a. Find an expression for ω_min
b. Evaluateω_minin rpm for a 65 g ball tied to a 1.0-m-long string.

Short Answer

Expert verified

a) The expression $ω_{m i n} = \sqrt{\frac{g}{r}}$

b) The value of ω_min = 30 rpm for the given ball.

Step by step solution

Part(a) Step 1: Given Information

mass = m
string length= L.

Part(a) Step 2 : Explanation

First draw a free body diagram as below then equate forces

At top point equate vertical force, we get

$T = (\frac{m v^{2}}{r}) - m g$

To avoid slack this force must be greater than or equal to zero

That means

$\frac{m v^{2}}{r} - m g \geq 0 \frac{v^{2}}{r} - g \geq 0 (C a n c e l i n g m) v \geq \sqrt{r g}$

Substitute v=ωr

$ω r \geq \sqrt{r g} ω \geq \sqrt{\frac{g}{r}}$

$ω_{m i n} = \sqrt{\frac{g}{r}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

Part(b) Step 1 : Given information

mass = 65 g ball

Length of string =1.0-m

Part(b) Step 2 : Explanation

Substitute the values in equation (1), we get

$ω_{m i n} = \sqrt{\frac{g}{r}} = \sqrt{\frac{9.8 m / s^{2}}{1 m}} = 3.132 r a d / s e c$

Convert into rpm

$ω = (3.132 r a d / s) (\frac{1}{2 π r a d}) (\frac{60 s e c}{1 m i n}) = 29.908 r p m \approx 30 r p m .$

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Speed m/sec	Scale Reading N
5	599
10	625
15	674
20	756
25	834

Short Answer

Step by step solution

Part(a) Step 1: Given Information

Part(a) Step 2 : Explanation

Part(b) Step 1 : Given information

Part(b) Step 2 : Explanation

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