Question

An athlete whirls an 8.54 kg hammer tied
to the end of a 1.4 m chain in a horizontal
circle. The hammer moves at the rate of
0.916 rev/s.
What is the centripetal acceleration of the
hammer? Assume his arm length is included
in the length given for the chain.
Answer in units of m/s

Answer 1

F/m = v^2/r = a

v = 2 pi r x 0.916 circumference travelled 0.916 times per second

answer a is in m/s^2 not m/s

Answer 2

im still not understanding the answer. sorry

Answer 3

@douglaswinslowcooper
why don't we use \(\alpha = \omega^2r\) since we have \(\omega = 0.916~~ rev/s\)

Answer 4

@ybarrap

Answer 5

When whirling around the hammer, the hammer has a constant speed, but because direction is changing, their will be an acceleration towards the center of the circle called the centripetal acceleration \(a_n\). The tangential acceleration \(a_t\), in the same direction of velocity and of motion, is the linear acceleration:

$$
\mathbf{v} (t) =v(t) \frac {\mathbf{v}(t)}{v(t)} = v(t) \mathbf{u}_\mathrm{t}(t) , \\
\begin{alignat}{3}\mathbf{a} & = \frac{\mathrm{d} \mathbf{v}}{\mathrm{d}t} \\ & = \frac{\mathrm{d}v }{\mathrm{d}t} \mathbf{u}_\mathrm{t} +v(t)\frac{d \mathbf{u}_\mathrm{t}}{dt} \\ & = \frac{\mathrm{d}v }{\mathrm{d}t} \mathbf{u}_\mathrm{t}+ \frac{v^2}{r}\mathbf{u}_\mathrm{n}\ , \\\end{alignat}
$$

Centripetal acceleration (in the normal direction) is
$$
a=\cfrac{v^2}{2}\\
\vec a=-\omega^2\vec R\\
|\vec a|=\omega^2R
$$

\(\omega\) is angular velocity in radians per second: \(\omega =\cfrac{2\pi }{T}\), where \(T\) is the period (the time to make 1 complete revolution, which we are given to be \(\cfrac{1}{0.916}\) seconds per revolution.

With the given information we can now compute the centripetal acceleration:
$$
|\vec a|=\left (2\pi \times 0.916\right )^2R\\
=\left (2\pi \times 0.916\right )^2\times1.4\\
=46.3~m/s^2
$$
towards the center.

Answer 6

Yes. v^2/r = r omega^2, so can use whichever.
Fine explanation by @ybarrap.

Answer 7

Note that $$\omega \ne 0.916~~ rev/s=\cfrac{1}{T}$$

Answer 8

thanks guys. if you can can you tell me th tension of of the chain ?

Answer 9

Well, the tension in the chain, the force it provides, is a little tricky to determine. The tension due to the centripetal acceleration is just T = F = m v^2/r, mass times the centripetal acceleration you calculated. In fact, gravity will be giving a downward acceleration also, so the chain will not be horizontal, even though the circle is. You can solve this, but the radius of curvature will not be the length of the chain. I suspect this is more complicated than your teacher intended you to solve.