Missing answers - sanity check needed

**Wonkihead** · August 10th 2008, 07:51 AM

I have a worksheet with answers missing. The questions are simple uniform motion in a circle ones but I would like to run through two of them here and just have you more experienced folks confirm my answers, if you wouldn't mind.

Question 1

Find the speed of a particle of dust, 5cm from the center of a disc which is revolving at 45rpm. Also find the acceleration of the particle. Answers to 2 decimal places.

Answer:

$ \omega = 45\text{ rpm} = \frac{45(360)}{60} \text{ deg s}^{-1} = \frac{45(360)(\pi)}{60(180)} \text{ rad s}^{-1} = \frac{3}{2}\pi \text{ rad s}^{-1} $

For the speed of the particle

$ v = r\omega = 0.05\bigg(\frac{3}{2}\pi\bigg) = { \color{blue}0.34 \text{ ms}^{-1} } \text{ (2 d.p.)} $

For the acceleration of the particle

$ a = r\omega^2 = 0.05\bigg(\frac{3}{2}\pi\bigg)^2 = {\color{blue}1.11 \text{ ms}^{-2} } \text{ (2 d.p.)} $

Question 2

A particle P of mass 20 grams is connected by a taut inextensible string of length 2m to a fixed point O of a smooth horizontal table on which the particle rests. The string will break if the tension exceeds 7.84N. Calculate the greatest possible speed at which P can describe a circle of 2m about O.

Answer:

$\text{mass, } m = 0.02 \text{ kg}$

For a maximum string tension of 7.84N

$F = ma$
$7.84 = 0.02a$
$a = 392 \text{ ms}^{-2}$

For the speed

$v^2 = ra^2$
$v = \sqrt{ 2(392)^2 }$
$v = {\color{blue}554.37 \text{ ms}^{-1} } \text{ (2 d.p.)}$

Thanks!

**topsquark** · August 10th 2008, 03:31 PM

Originally Posted by Wonkihead

I have a worksheet with answers missing. The questions are simple uniform motion in a circle ones but I would like to run through two of them here and just have you more experienced folks confirm my answers, if you wouldn't mind.

Question 1

Find the speed of a particle of dust, 5cm from the center of a disc which is revolving at 45rpm. Also find the acceleration of the particle. Answers to 2 decimal places.

Answer:

$ \omega = 45\text{ rpm} = \frac{45(360)}{60} \text{ deg s}^{-1} = \frac{45(360)(\pi)}{60(180)} \text{ rad s}^{-1} = \frac{3}{2}\pi \text{ rad s}^{-1} $

For the speed of the particle

$ v = r\omega = 0.05\bigg(\frac{3}{2}\pi\bigg) = { \color{blue}0.34 \text{ ms}^{-1} } \text{ (2 d.p.)} $

For the acceleration of the particle

$ a = r\omega^2 = 0.05\bigg(\frac{3}{2}\pi\bigg)^2 = {\color{blue}1.11 \text{ ms}^{-2} } \text{ (2 d.p.)} $

Question 2

A particle P of mass 20 grams is connected by a taut inextensible string of length 2m to a fixed point O of a smooth horizontal table on which the particle rests. The string will break if the tension exceeds 7.84N. Calculate the greatest possible speed at which P can describe a circle of 2m about O.

Answer:

$\text{mass, } m = 0.02 \text{ kg}$

For a maximum string tension of 7.84N

$F = ma$
$7.84 = 0.02a$
$a = 392 \text{ ms}^{-2}$

For the speed

$v^2 = ra^2$
$v = \sqrt{ 2(392)^2 }$
$v = {\color{blue}554.37 \text{ ms}^{-1} } \text{ (2 d.p.)}$

Thanks!

I didn't check the numbers, but the Physics is correct.

Just as a note for problem 2 I would specify that both the force and acceleration are "centripetal" for the sake of clarity.

-Dan

**Wonkihead** · August 10th 2008, 05:49 PM

Originally Posted by topsquark

I didn't check the numbers, but the Physics is correct.

Just as a note for problem 2 I would specify that both the force and acceleration are "centripetal" for the sake of clarity.

-Dan

Thanks for taking a look Dan.

I have a diagram in my working which I didn't include in this thread which shows the tension acting along the string from particle P towards O. My question is, given Newton's Third Law (equal and opposite forces), should tension act in equally along both $\overrightarrow{PO}$ and $\overrightarrow{OP}$ ?

This should probably be obvious but it is something I've never been absolutely clear on and never thought to ask...

It seems logical to me that it would, but having never had confirmation one way or another, I'm not sure either way.

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