- #1

VenomHowell15

- 14

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http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype11/prob03_pendulum.gif

There's the question I'm given. I cannot get terribly far with this one, despite being able to ace all the other questions on the assignment... Nonetheless, that is moot.

There are 3 unknowns here for this problem... Fortunately, one of the unknowns can be written in terms of the other one. The radius of the lower circle can be written as (r)(sin(theta)) which is in turn (1.490)(sin(theta)). I believe tension can be written as T=mg/(cos(theta)), but I'm not entirely sure.

I've attempted to do (mv^2)/(r(sin(theta)) = (mg)/(cos(theta)) and work it through to get ((v^2)/(gr)) = (tan(theta)), but it's not giving me the correct answer.

I'm kind of stuck, not sure where I'm going wrong with this.

**A mass of 8.700 kg is suspended from a 1.490 m long string. It revolves in a horizontal circle.**

The tangential speed of the mass is 3.755 m/s. Calculate the angle between the string and the vertical (in degrees).The tangential speed of the mass is 3.755 m/s. Calculate the angle between the string and the vertical (in degrees).

There's the question I'm given. I cannot get terribly far with this one, despite being able to ace all the other questions on the assignment... Nonetheless, that is moot.

There are 3 unknowns here for this problem... Fortunately, one of the unknowns can be written in terms of the other one. The radius of the lower circle can be written as (r)(sin(theta)) which is in turn (1.490)(sin(theta)). I believe tension can be written as T=mg/(cos(theta)), but I'm not entirely sure.

I've attempted to do (mv^2)/(r(sin(theta)) = (mg)/(cos(theta)) and work it through to get ((v^2)/(gr)) = (tan(theta)), but it's not giving me the correct answer.

I'm kind of stuck, not sure where I'm going wrong with this.

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