1. A person starts from rest at the top of a large, frictionless, spherical surf

Question

1. A person starts from rest at the top of a large, frictionless, spherical surface, and slides into the water below (see the drawing). At what angle ? does the person leave the surface? (Hint: When the person leaves the surface, the normal force is zero.)

1. A person starts from rest at the top of a large, frictionless, spherical surface, and slides into the water below (see the drawing). At what angle ? does the person leave the surface? (Hint: When the person leaves the surface, the normal force is zero.) Degree 6-42.gif 2. A 0.60 kg basketball is dropped out of the window that is 5.5 m above the ground. The ball is caught by a person whose hands are 1.9 m above the ground. How much work is done on the ball by its weight?

Explanation / Answer

!.As the person starts moving down the sphere, it is describing circular motion with increasing tangential velocity. This velocity is given by

1/2 m V^2 + mg (h - y) = mgh

where mgh is the total energy (constant) and y is the vertical distance from the top of the sphere to the point where the person is at the moment. Solving for V we get

V^2 = 2gy

Now, y is related to the curvature of the sphere as

y = R - R cos(angle)

where this angle is subtended between the point from the center of the sphere to the surface of it (where the person is) and the vertical. Before the person leaves the surface, this person is describing a circular motion with a tangential velocity V. The centripetal acceleration is then given by

ac = V^2 / R = 2gy / R = 2g (R-Rcos(angle)) / R
ac = 2g (1-cos(angle))

The centripetal force is then

Fc = m ac = 2mg (1-cos(angle))

This centripetal force is produced by the component of the person's weight that goes straight to the center of the sphere, this is

Fw = mg cos(angle)

At the start of the motion and for some distance traveled by the person Fw > Fc. As the person goes along, Fw decreases and Fc increases. The moment the component of the person's weight towards the center of the sphere is less than the centripetal force required to keep the circular motion is when the person "takes off" from the surface. This happens at a particular angle=angle_o where

Fc = Fw

mg cos(angle_o) = 2mg (1-cos(angle_o))

cos(angle_o) = 2(1-cos(angle_o))

cos(angle_o) = 2/3

thus

angle_o = 48 degrees

and that is your answer. Now, I don't see any drawing so you need to be careful with my definition of angle_o. This angle is subtended by a line that goes from the center of the sphere to the point at which the person takes off and a vertical line.

2.W = F d (force x distance) F = mg , F = (0.6 kg)(9.8 m/s^2) = 5.88 N
W = (5.88 N)*(5.5 - 1.9)
W = 21.168 J

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