0

A small mass is suspended by a light rod from a pivot P. The mass moves with constant speed in a horizontal circle. The rod has length 1 metre and makes an angle of $30^\circ$ with the vertical. Assume $g=9.8ms^{-2}$

I solved part $A$ and $B$ calculating that the mass takes $1.87$seconds to complete one revolution. I also solved that the speed of the mass is about $1.68m/s$

However part $C$ says, if the speed of mass is doubled, show that the rod will make an angle of $54^\circ 44 '$ with the vertical.

THIS means $3.36m/s$. But how would you calculate this. You know equations such as velocity = radius $\times \omega$ . However, radius changes because you are going faster. Angular velocity also changes because you are going faster.

I have also considered speed $= \frac{\text{distance}}{\text{time}}$

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
D.Ronald
  • 540
  • What is the centripetal acceleration (and hence force) required to maintain the circular motion? – amd Oct 04 '17 at 01:06
  • well, this is the same as you starting question, but backwards - instead of being given the angle and looking for everything else, you are now given the speed, and looking for everything else. Just rearrange your equations with which you solved A and B, to make the angle the "principal" unknown – Nick Pavlov Oct 04 '17 at 11:48

0 Answers0