A particle moves along a circle of radius $\frac{20}{\pi}$ with constant tangential acceleration.

Question

If the speed of the particle is $80\frac{\text{m}}{\text{s}}$ at the end of second revolution. Find the tangential acceleration (object starts from rest)

$V=R\omega$

$\omega =4\pi$ rad/s

Now $$\omega^2=2\alpha\theta$$ $$16\pi^2=8\pi\alpha$$ $$\alpha=2\pi $$

Then $$a=R\alpha$$ $$a=40m/s^2$$

But the answer is $80.$ What’s going wrong?

To be clear, I have a perfectly valid solution with me that is convincing enough to get 80m/s^2. I just want to know what’s wrong with this

Upon request, here is the solution

Answer 1

Your answer of $a = 40$ m/s$^2$ is correct. Assuming $2$ revolutions, the proposed solution has an arithmetic mistake. The first line is correct, but the second line is wrong. It should be \begin{align} a &= \dfrac{v^2}{2d} = \dfrac{v^2}{2 (2 \cdot (2 \pi R))} = 40 \end{align} First equality is by rearranging $v = \sqrt{2ad}$. The second is by substituting $d = 2(2 \pi R)$, which is the distance travelled at the end of two revolutions and the third is by substituting $v=80$ at the end of two cycles, $R= 20/\pi$.

So, the mistake is that the solution is missing an extra factor of $2$ in the denominator.

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(of course, insert units everywhere to be formally correct)

Answer 2

Your solution seems correct, and the book's solution is wrong. First of all, $d = R\theta = 4\pi R.$ Look at the first 2 equalities after the "$\therefore$":

$a = \frac{v^2}{ad} = \frac{v^2}{2(2\pi R)}$

The first equality has a typo; it should be $\frac{v^2 }{2d}$. Assuming this, $\frac{v^2}{2d} = \frac{v^2}{2(2\pi R)}$ so $d = 2\pi R \neq 4\pi R, \quad\Rightarrow\!\Leftarrow.$