Circular Motion: Solving for Tension and Speed of a Bead on a String

AI Thread Summary
A 100g bead slides along a 0.8m string attached to a vertical pole, with the ends 0.4m apart. When the pole rotates, the string segment BC becomes horizontal at 0.3m. The tension in the string can be derived from the forces acting on the bead, leading to the equations Tsin(θ) = mg and Tcos(θ) + T = mv²/r. The calculations yield the tension T as 1.25mg and the speed of the bead at B as 66.67 m/s. The discussion emphasizes the relationship between tension, centripetal force, and gravitational force in circular motion.
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A 100gm bead is free to slide along a 0.8m long piece of string ABC. The ends of the string are attached to a vertical pole at A and C which are 0.4m apart. When the pole is rotated about its axis, BC becomes horizontal and equal to 0.3m.
(a) Find the tension in the string
(b) Find the speed of the bead at B.

I don't know where to start with this problem, could someone please help?

Thanks
 
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flower76 said:
A 100gm bead is free to slide along a 0.8m long piece of string ABC. The ends of the string are attached to a vertical pole at A and C which are 0.4m apart. When the pole is rotated about its axis, BC becomes horizontal and equal to 0.3m.
(a) Find the tension in the string
(b) Find the speed of the bead at B.

I don't know where to start with this problem, could someone please help?
The tension in the string provides two forces on the bead. The forces have the same magnitudes but different directions. The horizontal component of these forces provides the centripetal force and the difference in vertical components provides the normal force (mg)

Tsin\theta + T = T + .3T/5 = 1.6T = mv^2/r

Tcos\theta = .4T/.5 = .8T = mg

So: 2g = v^2/r where r = .3 m
and: T = 1.25mg

AM
 
The answer is 66.67 ms^1
contact me for mre info.

a hint- T sin ά=mg

T cosά+T=mv*v/r
 
Andrew Mason said:
The tension in the string provides two forces on the bead. The forces have the same magnitudes but different directions. The horizontal component of these forces provides the centripetal force and the difference in vertical components provides the normal force (mg)

Tsin\theta + T = T + .3T/5 = 1.6T = mv^2/r

Tcos\theta = .4T/.5 = .8T = mg

So: 2g = v^2/r where r = .3 m
and: T = 1.25mg

AM
The answer is 66.67 ms^1


T sin ά=mg

T cosά+T=mv*v/r
If not for a care less mistake this should be correct.
 
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