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We have learned that Torque is equal to a force that is perpendicular to a radius (displacement); however, I just cannot grasp one of the study questions we received:

A hammer thrower accelerates the hammer (mass = 7.30 kg) from rest within four full turns (revolutions) and releases it at a speed of 30.0 m/s. Assuming a uniform rate of increase in angular velocity and a radius of 1.50 m, calculate the following answers.

(a)the angular acceleration: 7.96 rad/s2

(b) the (linear) tangential acceleration: 11.94 m/s2

(c) the centripetal acceleration just before release: 600 m/s2

(d) the net force being exerted on the hammer by the athlete just before release: 4380 N

(e) the angle of this force with respect to the radius of the circular motion is___

So basically, I have answered the angular acceleration, tangential acceleration, centripetal acceleration and the net force, but I cannot determine the angle. Thanks!

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When do you want the angle? As the hammer spins up, the centripetal force must increase while the tangential force remains the same, so the angle decreases with time. It's a little curious where the tangential force comes from, if the hammer is moving in a circle connected to the center by a radius. (Try watching some videos of people tossing hammer. They actually move forward through the throw area while spinning up.) –  Mark Eichenlaub Feb 2 '11 at 21:10
    
I believe that the tangential acceleration referred to the linear displacement after the hammer had been released. The previous question asks for the net force exerted on the hammer right before release, so the angle must be right before release. Would posting the previous Qs and As help? Thanks! –  MetroGnome Feb 2 '11 at 21:16
    
Yes, definitely post the previous parts of the question because they're necessary to make sense of part (e). Also tell us what you've tried to do with this part. You need to put more effort into it than just saying "I can't solve this question." –  David Z Feb 2 '11 at 21:33
    
@Metro First of all, this is not a discussion forum. You ask question and get answer, fin. You can express and share your feelings/thoughts on chat. If you want to know something more about how this site works, check out the FAQ or ask on meta. Have a nice flight! –  mbq Feb 2 '11 at 21:38
    
Thanks for all the feedback guys! filling in additional info now. –  MetroGnome Feb 2 '11 at 21:40

3 Answers 3

up vote 2 down vote accepted

You know the tangential and centripetal acceleration so it's good to use these directions as base. And you know that the direction of the force is the same as the acceleration. Meaning, you have the direction of the force in the plane the hammer is rotating $\vec{v} = (-600, 11.94)$. In this same plane you know the direction of the radius: it's centripetal $\vec{u} = (1,0)$. Now you just calculate the angle between $\vec{v}$ and $\vec{u}$.

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Thanks! That fixed it –  MetroGnome Feb 13 '11 at 20:09

The force acted by the thrower on the hammer has two components: centripetal and tangential. What you actually computed in point (d) was only the centripetal component, related to the centripetal acceleration. If you also compute the tangential component of the force, you'll be able to calculate:

  1. the correct net force, using Pythagoras' theorem
  2. the angle at which this force is acting, using some basic trigonometry (a sketch of the situation seen from above will help you determine the correct relation to use)
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For linear acceleration you know you find distance by
Distance = 0.5 x acceleration x time^2

That would solve your problem for the linear case. Try looking around for something similar in the rotational case. There are a bunch of ways of doing the problem even starting with my hint.

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protected by Qmechanic Jun 30 '13 at 18:52

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