Question

Tony Hawk is on his skateboard initially at rest, and drops into a half pipe of radius 3 m at the rim as shown in the figure. The instant he reaches the bottom, the pipe exerts a normal force of Fn. Later that day, he drops in the same way into a half pipe of radius 6 m. When he reaches the bottom of that pipe, it exerts a normal force of

a) 2Fn b) 4Fn c) √ 2Fn d) Fn e) 1/2 Fn

Please explain ur answer

Thanks

Answer 1

At the botttom, writing the force equation,

N - mg = mv^2/R

where N = normal force acting on him towards up

v = speed at the bottom

R = radius of the circular path

So, N = mg + mv^2/R ----------- (1)

Using energy conservation, the PE at the top of the pipe = mgR <---- height ,h = R

Now, after reaching the bottom, the PE is converted to KE

So, KE = 0.5*mv^2 = mgR

So, mv^2 = 2mgR

Plugging this in equation 1

N = mg + (2mgR)/R = mg + 2mg = 3mg

We can clearly see that the normal force does not depend on the radius of the pipe (R)

So, the normal force will be same as before

So, answer is d) Fn