A girl from a discus team throws a discus and her arms make a circle with radius

Question

A girl from a discus team throws a discus and her arms make a circle with radius of 1.2m around her central axis of rotation. She releases the discus perpendicularly from her axis of rotation and at a height of 1.5m. In her wind up to throw the discus, she rotates at a rate of .8 rotations per second. So if she releases the discus tangently to her axis of rotation and horizontally, how long will the discus be in the air before it lands? How far will it travel from the point at which she releases it?

With all the same data as above, if the girl can also run at 1 m/s while rotating in the direction of the throw at the time she releases the discus, How long will the discus be in the air and how far will it travel from the point at which she releases it?

Explanation / Answer

The first part is simple. Since she is releasing it horizontally, there is not intial y velocity, so it can be treated as a free fall problem to find the time. Use the kinetic equation

x = vot + (1/2)gt2

Where vo = 0 m/s, x = 1.5 m, and g = 9.8 m/s/s

Solve for t.

Next, we find the angular velocity in radians per second, so if she is moving .8 rotations per second, we can convert it by:
(.8 rotations / 1 sec)(2 radians / 1 rotation)

So that the angular velocity, , is 2*0.8 radians / second.

Now we want the linear velocity, and we know that angular and linear velocity are directly proportional by

v = * r

v = (2)(0.8)(1.2 m)

Since we know how long the disc is falling, found in part I, we know how long it will be travelling horizontally. Since

v = d/t, then

d = v*t

So plug in the found velocity and the time, and solve for d.

If she is running at 1 m/s, the only thing that will change is the last part, how far it will go horizontally. Instead of plugging in the velocity found by v = r, plug in that velocity PLUS 1 m/s and solve for d in the same manner.

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