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Circular Motion

1_ Particle P attached to light inextensible string which is 2m, the other end's attached to fixed point O. Particle P moves in a horizontal circle with constant speed, v, with the string taut and inclined at a constant angle of 20* to the vertical. Find v.

I've drawn the diagram with P having a tensions outwards to the point O at the top, the angle between the tension and the vertical is 20* (vertical being where a normal would have been if a surface was involved). So, I split components to get:

Tcos20=mg (vertically)
Tsin20=ma (horizontally) -> Tsin20=m(v^2/r) -> 2Tsin20=mv^2

Then I rearranged the vertical for T and put it into the horizontal. m's cancel. I'm left with:

(2gtan20)^1/2=v, v=2.67, but the answer beside the question is 1.56.

2_ A conical pendulum consists of a particle attached to the end of light inextensible string, length .8m, the particle moves in a horizontal circle and the system rotates at constant angular speed 4 rad/s; find the angle string makes with vertical.

I've drawn the diagram pretty much like the last one, except this time there's theta instead of a value for the angle and:

Tcos theta = mg (vertical)
Tsin theta = m r w^2 (horzontally, w being omega)
Rearranged vertical for T and put into horizontal and I get:
gtan theta = 0.8x4^2 -> theta = 52.5*

The answer says 40*

(edited 9 years ago)

Reply 1

Random1357

Original post by EdwardHeissinger

Right. Hey guys, I'm on a gap year and I've recently begun some revision for some exams I'm retaking and so I'm solving some mechanics problems I got off of another student but I'm sort of stumped and so decided to join this forum. Could someone help point out what I'm doing wrong or what to do right for these problems? It'd be really useful because I feel like I've forgotten everything!

1_ Particle P attached to light inextensible string which is 2m, the other end's attached to fixed point O. Particle P moves in a horizontal circle with constant speed, v, with the string taut and inclined at a constant angle of 20* to the vertical. Find v.

I've drawn the diagram with P having a tensions outwards to the point O at the top, the angle between the tension and the vertical is 20* (vertical being where a normal would have been if a surface was involved). So, I split components to get:

Tcos20=mg (vertically)
Tsin20=ma (horizontally) -> Tsin20=m(v^2/r) -> 2Tsin20=mv^2

Then I rearranged the vertical for T and put it into the horizontal. m's cancel. I'm left with:

(2gtan20)^1/2=v, v=2.67, but the answer beside the question is 1.56.

2_ A conical pendulum consists of a particle attached to the end of light inextensible string, length .8m, the particle moves in a horizontal circle and the system rotates at constant angular speed 4 rad/s; find the angle string makes with vertical.

I've drawn the diagram pretty much like the last one, except this time there's theta instead of a value for the angle and:

Tcos theta = mg (vertical)
Tsin theta = m r w^2 (horzontally, w being omega)
Rearranged vertical for T and put into horizontal and I get:
gtan theta = 0.8x4^2 -> theta = 52.5*

The answer says 40*

1) r is 2 sin (20)