A small lead ball, attached to a 1.70-m rope, is being whirled in a circle that

Question

A small lead ball, attached to a 1.70-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 0.85 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise?

The skateboarder in the drawing starts down the left side of the ramp with an initial speed of 5.2 m/s. Neglect nonconservative forces, such as friction and air resistance, and find the height h of the highest point reached by the skateboarder on the right side of the ramp.

Explanation / Answer

We will start by finding the speed at which the ball leaves the circle.

We know that the circle has a radius r=1.7m.

So its perimeter will be P=2*pi*r = 2*3.14*1.7=10.68m
Its given that the ball whirled 3 complete circle of 10.68m each second. So its tangential speed will be
v=3*P/1s = 3* 10.68m / 1s = 32 m/s

Now the problem converts into a constant acceleration problem. The ball go upward at 32 m/s(This is the tangential speed.) and decelerate at -9.81m/s^2(this is acceleration due to gravity).

We will calculate upto what height does the ball rise.

In upward motion, at the rate -9.81m/s^2 to get to 0 m/s it takes the time: v=u+at

t = (32m/s) / (9.81 m/s^2) = 3.26 s

Thus, it takes 3.26 s and average speed will be (32m/s + 0m/s) / 2 = 16 m/s

Travelling at 16m/s average for 3.27s makes a distance of : d = 16m/s*3.26s = 52.16 m

Total height is d + 0.85m = 52.16 m + 0.85m = 53.01m

Here all the kinetic energy will change into potential energy (no frictional losses).
KE = PE (energy conservation problem)
KE = 0.5 mv^2       (final KE is zero as v=0 at highest point)
PE = mgh              (initial PE is zero as on ground, height is zero.)
So, 0.5mv^2 = mgh
where g = 10m/s^2 and m is common on both sides.
So 0.5v^2 = 10h
So h = 0.5 x 5.2^2 /10 = 1.352m

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