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I must respond to this about the discuss: (A) the direction of the centripetal exerted...

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quddoos | (Level 1) Valedictorian

Posted September 8, 2013 at 11:30 AM via web

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I must respond to this about the discuss: (A) the direction of the centripetal exerted on the force.

(B) the direction in which the thrower experiences a centrifugal force on the discuss.

(C) the direction in which the discuss would go when the discuss thrower releases it.

(D) why the discuss would go in the direction it does (which of Newton's Law would it obey)?

Discuss throwing is one of the oldest individual sport. The ancient Greeks included it in their Olympic games. In this sport the thrower holds in the palm of his or her hand a flat plate or disk, whirls around in a circular motion and lets the disk go. Assume that you are an outside observer. Draw and label arrows on the illustration to indicate the responses to the above.

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llltkl | College Teacher | (Level 3) Valedictorian

Posted September 8, 2013 at 4:04 PM (Answer #1)

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In the case of discus throwing, the force that acts as the centripetal force is that exerted by the hand of the thrower onto the discus (see the attached image).

The disc, however experiences a centrifugal force (amounting to `(mv^2)/r` ), which helps the thrower keep it "locked" into his throwing hand.

Once the disc is released, it would go in a tangential direction from the point of release.

Angular momentum is transferred from one body part to another. In throwing, Newton’s 3rd law (Law of reaction) is applied to transfer angular movement from one body part to another. A right hand thrower will block the left side of the body immediately before delivery in an attempt to transfer angular momentum of the body to linear movement of the discus. While the left side of the body is blocked (state of inertia), the right hand arm will accelerate to prepare for the final delivery of the discus.

The acceleration of the disc after release, on the other hand, can be otained from  Newton’s 2nd law of motion as:

`F=m*a=(mv^2)/r` 

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