In an experiment, a 0.028 kg rubber stopper is attached to one end of a string. A student whirls the stopper overhead in a horizontal circle with a radius of 1 meter.The stopper completes 10...

In an experiment, a 0.028 kg rubber stopper is attached to one end of a string. A student whirls the stopper overhead in a horizontal circle with a radius of 1 meter.

The stopper completes 10 revolutions in 10 seconds. What is the speed of the whirling stopper?

Asked on by acaputi

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bandmanjoe's profile pic

bandmanjoe | Middle School Teacher | (Level 2) Senior Educator

Posted on

What we need here to calculate the speed is the distance around the circle that is being created by the whirling stopper.  The mathematical formula "circumference equals pi times diameter"  would come in handy in calculating the distance the stopper is traveling in one revolution.  Pi is 3.14, rounded off, and the diameter would be twice the radius, which would be 2 meters.  So, if we substitute those values into the formula, it would look like this:

circumference = 3.14 x 2 meters

circumference = 6.28 meters

So the distance the stopper travels in one revolution is 6.28 meters.  Since it accomplished 10 revolutions in 10 seconds, the easy math there is one revolution takes 1 second.  So the stopper is traveling at a speed of 6.28 meters per second.  Speed is distance traveled over time, so 6.28 meters per second is the speed at which the stopper is moving.

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sammyrafeeh's profile pic

sammyrafeeh | Student, Undergraduate | eNotes Newbie

Posted on

f= 1 Hz (1 revolution every second)

r= 1 m

m= 0.028 kg

 

Use the equation of centripetal force is

Fc= m4∏2r2f2

Fc= 1.104 N

Now rearrange the equation to find the speed(v)

Fc= mv2/r

v= √Fcr/m

v= √ [(1.104 kg.m/s2)(1 m) / (0.028 kg)]

v= 6.28 m/s

So the speed of whirling stopper is 6.28 m/s 

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