Calculations for a Hammer Throw Event

[knapklara]

Homework Statement

Hammer thower spins at circular frequency of 0.5 s-1 at one point and with constant radial acceleration in the same direction α = 2 rad/s-2. What angular velocity ω will he have after anoter 1.5 twist? What tangential velocity will the hammer have when the distance from rotation axis is d = 1.5 m?

Relevant Equations

ω = 2π/t0= 2πv
α = const.
ω = αt + v0
d = r ?

I'm already confused by the terms circular, angular and radial. I've read the definitions but I'm still confused. What is the difference? Is ω the right symbol for frequency that I used?
Any tips much appreciated!

 

[kuruman]

Your use of symbols is confusing. Don't use v unless you intend to denote velocity or speed. I would write
$\omega = \omega_0+\alpha~t$ for the angular speed. It is related to the frequency $f$ by $\omega=2\pi~f$ and since the period $T$ is the inverse of the frequency ($f=1/T$) one can also write $\omega=2\pi/T$.

Circular means "going or that has to do with a circle". Circular motion means that something is going around in a circle. If you draw a fixed radius on the circle, the instantaneous position of the object will form an angle relative to the fixed radius. Because this angle is changing with time, the rate at which the angle changes with respect to time is called the angular velocity and is denoted by $\omega$.

"Radial" means in a direction directly away from the center of the circle. An object that moves in a circle of fixed radius has zero radial velocity. However it has centripetal acceleration $a_c=v^2/R$ which is directed not away but towards the center of the circle. Another way of saying this is, the radial component of the acceleration is $a=-v^2/R$.

I hope this clarifies things a bit.

 

[BvU]

What is the difference?

The difference between what and what ?
Google gets you plenty places with angular motion formulas, but some (this one on unform circular motion) are clearer than others. (basics, Wikipedia, Hyperphysics, )

We can't attempt to write better textbooks here on PF, so you 'll have to search for something you like.

Once around a circle is $2\pi$ radians. If the motion is uniform (constant speed $\vec v = \vec \omega \times \vec r\$), one revolution/second, you have $\omega = 2\pi$ rad/s

It is confusing to write 'ω = 2π/t₀' (but correct if t₀ is the time for one revolution in uniform circular motion), and it is wrong to write 'ω = 2πv' --- check the dimensions (but $\omega = 2\pi\nu$ is again right for uniform circular motion; a subtle difference beteween $v$ (meter/s) and $\nu$ (1/s) ).

And 'ω = αt + v₀' is also wrong: v₀ has the wrong dimension. (again: $\omega = \omega_0+\alpha t$ is OK).In your exercise you have constant angular acceleration, so the angular analog for the SUVAT equation for linear motion with constant acceleration (here $s = v_it +{1\over 2} at^2 \$, and -- more complete -- here : $x = x_0 + v_0 t + (1/2) a t^2$ ) applies:
$$\theta = \theta_0 + \omega_0 t + {1\over 2} \alpha t^2 \tag 1$$and now we have to link that to the situation in your exercise.

'At one point' we can state that the angle at that point is $\theta_0$.
'spins at circular frequency of 0.5 s-1' is supposed to give us $\omega_0$

I have a problem there: The angular velocity can be 0.5 rad/s, or the frequency can be 0.5 Hz, i.e. half a revolution per second.​

A little googling helps me out with my personal problem: the term circular frequency a.k.a. angular frequency that I find confusing refers to omega.​

So we can state $\omega_0 = 0.5$ rad/s.

Then: 'What angular velocity ω will he have after another 1.5 twist?' sets our $\theta-\theta_0$ in $(1)$ to $3 \pi$ radians.

That means we now have $$3\pi = \omega_0 t + {1\over 2} \alpha t^2 $$ which we can solve for $t$.
Once we have $t$ the analog of $v = v_0 + at\$ helps us find the $\omega(t)$ the exercise wants.

And the $v = \omega r$ gets the tangential velocity.

[edit] It took me some time to collect all this and type it up. @kuruman was much faster, but perhaps the combination of the two replies is even better than each of them on its own

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[haruspex]

Circular means "going or that has to do with a circle".

"Going in, or having to do with, a circle"

"Radial" means in a direction directly away from the center

Or directly towards the centre; but generally the positive direction is taken to be away.

As has been mentioned, in your $\omega=2\pi/t_0$ equation, $t_0$ is presumably the period, which is usually written T. But that is only valid for a constant rotation rate. Here there is an acceleration, so there is no "period". Rather, you are given an initial instantaneous rotation frequency of $0.5s^{-1}$, i.e. one complete turn every 2 secs if there were no acceleration. So $\omega_0=2\pi\frac 12=\pi$ radians/sec.

 

[knapklara]

OK, so after a little thinking I came up with this. Thank you everyone for your input!

 

[kuruman]

The equation for the time to do the 1.5 twist assumes that the hammer starts from rest. We are told that, at the start of the 1.5 twist, the hammer "spins at circular frequency of 0.5 s-1" and has constant angular acceleration α = 2 rad/s². We are not told how it started and how it got to that point.

If you want to find the time, you have to solve the quadratic equation that you wrote down $\varphi=\omega_0~t+\frac{1}{2}\alpha~t^2$. Alternatively, you can bypass the time calculation and use the equation that relates the angular acceleration, angular displacement and the squares of the angular speeds.

 

[haruspex]

OK, so after a little thinking I came up with this. Thank you everyone for your input! ,

Please do not post working as an image. It makes it hard for us to pick out and comment on specific steps, and sometimes handwriting is unclear. Take the trouble to type it in. Use LaTeX, or the superscript/subscript controls ($X^1, X_1$) above the text panel. Or failing that, just type ^ in front of an exponent and _ in front of a subscript: v_0=0.5s^(-1).

I think your first equation is $v_0=1.5s^{-1}$, or it could be $\nu_0=1.5s^{-1}$. For clarity, it would be better to use $\omega_0$ for the initial angular frequency.

I don't understand your fourth equation, $v=2s^{-1}$. What is this v and what given value is 2 per second?

Your fifth equation is ok, but it would be better with $\omega_0$ instead of just $\omega$. It is the initial angular frequency.

As @kuruman points out, you failed to include $\omega_0$ in your sixth equation.

 

[BvU]

Rather, you are given an initial instantaneous rotation frequency of $0.5$ s⁻¹, i.e. one complete turn every 2 secs if there were no acceleration.

Does that mean you disagree with post #3

'spins at circular frequency of 0.5 s-1' is supposed to give us $\omega_0$

I have a problem there: The angular velocity can be 0.5 rad/s, or the frequency can be 0.5 Hz, i.e. half a revolution per second. A little googling helps me out with my personal problem: the term circular frequency a.k.a. angular frequency that I find confusing refers to omega.
So we can state $\omega_0=0.5$ rad/s.

 

[haruspex]

Does that mean you disagree with post #3

I did have a different view, but I am persuaded you are correct.

Edited - thanks.

 

[BvU]

For the record: I went the same path ...

And I admit I have a hard time trusting wikipedia

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