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

[flower76]

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

 

[Andrew Mason]

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

 

[kusal]

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

a hint- T sin ά=mg

T cosά+T=mv*v/r

 

[kusal]

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.