Torque homework 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!
 A: 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 $\vec{u}$.
A: 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.
A: 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:


*

*the correct net force, using Pythagoras' theorem

*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)

