How far (measured along an arc in radians) from the catcher does the ball wind up?

Question

Two men are standing on a merry-go-round which is turning at a constant angular speed. One man is standing very near the center of the merry-go-round, the oher is standing very near the edge. the man at the center is trying to throw a baseball directly to his friend at the edge but the ball always winds up far away from target. The angular speed of merry-go-round is .1 radians/second and it is 12 meters in diameter. The baseball is thrown with a velocity of 13m/s exactly horizontally...how far from the catcher does the ball wiind up.


Also-- the catcher perceives that a force is acting on the ball to skew its path. This is a "fictitious force." What size force would be needed to cause the motion observed by the catcher if it were real?

Homework problem that I have benn givent the answer but I do not understand how it is done....if anyone can explain it to me step by step...thanks

Someone asked the question in class of what the mass was and we were told that you do not need it to figure out the problem....that's all I know...unless you just used m as a variable without an exact number

Ok j you are helping a lot but I still have one question...u say you used pythagoream but I do not understand why and how....that is where you are losing me

Answer 1

Sine the man throwing the ball and the ball are near the edge of the merry-go-round they have a tangential speed equal to w*r let's call the tangential speed s
The ball is thrown at 13 m/s (v) horizontally directly at the catcher but travels tangentially at .6 m/s. w*r

I will assume that the distance we want to calculate is the distance from the ball on an arc that is subtended by the thrower, has the same radius as the merry-go-round, and intersects the ball when the catcher should have caught the ball on a non-rotating frame, and the catcher at the center.

The flight time is
t=r/v

so the distance away from the catcher
can be calculated using pythagorean
d=sqrt(l^2+s^2*t^2)
where l is the distance the intersection of a perpendicular line of the location of the ball to the radius between the two men
s^2*t^2=w^2*r^2*r^2/v^2
=.077 m
l=r-sqrt(r^2-s^2^t^2)
=6-sqrt(36-.077)
=.006 m

Here's a diagram of my solution
http://i142.photobucket.com/albums/r88/odu83/merrygoroundba.jpg


d=sqrt(.006^2+.077)
=.28 m
about a foot


The fictitious force is also known as the Coriolis effect
a=2*w*v which is a formula for the special case of a merry-go-round

the force is 2*m*w*v
or
2*m*.1*13

j

Answer 2

The guy in the centre tries to throw it directly to his friend , it travels 12/2 metres in 12/13 seconds but in that time, the merry-go-round has turned through 0.1*12/13/2 radians so it misses his friend who is a very thin person) by this many radians.

You can't determine what the fictitious force appears to be unless you know the mass of the baseball, what the guy in the centre seems to see is an acceleration of the ball sideways such that it moves 12*.1/2 meters away from his friend in 12/13/2 seconds so the apparent acceleration is 2*12*0.1/2/(12/13/2)^2 = 12/2*0.1*13^2 or about 3.1/2m/s/s

Sorry about that - I used 12 as the radius at first. Then messed up on the time^2 bit.

Answer 3

The ball is thrown at 13 m/s for a distance of 6 m. Therefore it takes 6m/13m/s = 6/13 seconds to get there.

In that time (6/13 seconds) the catcher - moving at .1 rad/sec will have moved .1 rad/sec x 6/13 seconds = 6/130 of a radian.

The merry-go-round is 12 m in diameter. Its circumference is 12m x 3.14. But the catcher only moves 6/130 of a radian. Since there are 2*3.14 radians in a circle. the catcher moves only 6/130 divided by 2*3.14 of the circumference.
(3/130*3.14)

Multiply this by the circumference (12 *3.14) and you get
36/130 m or about 0.277 m. or 11 inches.
If my arithmetic is correct.

Re the force problem, I think we need to know the mass of the ball. F=ma

Answer 4

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