Suppose a rock of mass 2.2 [kg] is attached to a string of length 1.00 [m] is
twirled around in a perfect circle as a speed of 2.5 [m/s]. Calculate the work
done on the mass by the tension force exerted on the rock by the string
during one revolution of the rock. Ignore the effect of gravitational forces.
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We need to deal with two aspects of this question: Work and Circular Motion.
- Work is equal to a force acting over a distance: W = Fd
- Circular motion is a lot like linear motion, but with the inclusion of the circle's radius, as well as the force keeping the body moving in a circle. This is the Centripetal Force, or `F_c` . The centripetal force always acts perpendicular to the linear motion. This allows it to change the direction, but not the speed, of the object, and keep it moving in a circular path.
The solution is relatively simple. In order to find the Work done, we need to know what the force is, and the distance that it acted over. Again: W = Fd.
To find the distance traveled by the rock, we look at its path. It traveled in a circle; a circle's circumference is the "distance", or length, of its outer edge. Circumference is calculated by 2pi times the radius. For a radius of 1m;
2pi (1) = 6.28m
The force is acting over a distance of 6.28m
The force itself may be a bit less intuitive; it's the centripetal force. This force is the exact same thing as tension.
We know that centripetal force is equal to `(mv^2)/(r)`
So the force equals `((2.2)(2.5^2))/(1)`
Which works out to 13.75N
Work = F(d)
Work = 13.75 x 6.28
Work = 86.35 Joules
Since we're limited by 2 significant figures, `W = 86J`
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