- #1

maviss

- 8

- 0

## Homework Statement

The problem involves a hammerthrow with a ball on the end of a cable being spun around and released by an athlete. The hammer is spun four times before release and is released at ground level at an angle 40 degrees to the horizontal.

mass of hammer: 7.678kg

length of cable: 1.562m

velocity of hammer upon release: 28.23m/s

In this section of the question, we are asked to find the tension force in the cable at the time of release.

## Homework Equations

F = ma

F(net) = mv^2/r

## The Attempt at a Solution

At first I thought that I could simply use trigonometry and the fact that the hammer is released 40 degrees from the horizontal to find the tension force. I drew a free body diagram of the weight and identified two forces - the tension force and the weight force. I found that the weight force was m x g = 7.678 x 9.8 = 75.24 N. Then I used cos (theta) = A/H and rearranged to find the tension force H (by subbing in theta = 40 and A = 75.24) to find that H = 98.22N.

When I thought about it more closely, I realized that the problem is one of circular motion and that the force exerted by the athlete is the only force acting towards the centre of the circle and that it must be the centripetal force in the system. So I used the equation F(net) = mv^2/r to find that Fc = (7.678 x 28.23^2)/1.562.

I got that Fc = 3917N.

So then I thought that Fc = T in this case because T also pulls the weight toward the centre of the circle, and that therefore T = 3917N.

Is the latter method correct? And what happens to the weight force I used originally, does it not matter when you're calculating the centripetal force?

## Homework Statement

The problem premise is the same as before, only this time we are asked to find the net torque that the athlete exerts on the hammer. We are asked to assume that the angular acceleration is constant, and that all of the mass of the hammer is found in the ball.

mass of hammer: 7.678kg

length of cable: 1.562m

velocity of hammer upon release: 28.23m/s

from section above, tension force: 3917N.

## Homework Equations

[tex]\omega[/tex] = v/r

2[tex]\alpha[/tex][tex]\delta[tex][tex]\theta[/tex] = ([tex]\omega[/tex]f)^2 - ([tex]\omega[/tex]i)^2)

a (tangential) = [tex]\alpha[/tex]r

[tex]\tau[/tex] = F x r x sin [tex]\Phi[/tex]

## The Attempt at a Solution

So again, at first glance, I think I got the wrong answer. I originally thought that because the axis of rotation is the athlete that the torque supplied by the athlete must be zero. But now I don't think that's right, is it?

Because there is an angular acceleration? And for there to be an angular acceleration, there must be torque. So I figured that I needed to find the force that was applied tangential to the circle, since the hammer will travel in this direction when it is released and the athlete must be applying a force tangential to the circle for there to be an angular acceleration - is this correct?

So, I found the angular acceleration using:

2[tex]\alpha[/tex]delta[tex]\theta[/tex] = ([tex]\omega[/tex]f)^2 - ([tex]\omega[/tex]i)^2)

The initial angular velocity is 0 because the athlete spins the hammer from rest. The final angular velocity must be the velocity/the radius, so: 28.23/1.562 = 18.07radians/s. Subbing in these values, and because the hammer is spun four times traveling a total distance of 8pi radians

I got:

2 x [tex]\alpha[/tex] x 8pi = 18.07^2 - 0

Therefore [tex]\alpha[/tex]= 6.498 radians per second^2

Then I used [tex]\alpha[/tex] to find the tangential acceleration by multiplying [tex]\alpha [/tex] by 1.562 to find at = 10.15.

Then I used this to find the force tangential to the circle according to F = ma.

F = 7.678 x 10.15 = 77.93N

And then I used this value to calculate the torque. The radius is the distance from the axis of rotation and the angle at which it is applied is perpendicular to the axis of rotation, so:

[tex]\tau[/tex] = 77.93 x 1.562 x sin (90)

= 121.7

Is this correct? Or have I just taken the question completely wrong? I'm really unsure about this part - I spent ages trying to figure it out and I don't know if I'm just overthinking it.

Any help and advice appreciated, thanks so much!