Pendulum problems - determine speed of the ball

[logglypop]

A ball is attached to a string with length of L. It swings in a horizontal circle, with a constant speed. The string makes an angle (theta) with the vertical, and T is the magnitude of the tension in the string.

1)Determine the Mass of the Ball.
2)Determine th Speed of the Ball.
3)Determine the Frequency of revolutions of the Ball.

1)
F=Tsin(theta)=ma
m= (Tcos theta)/g

2)
F= (mv^2)/a= Tsin(theta)
v^2= [ aTsin(theta) ] / m
v= square root of [ aTsin(theta) ] / m

3)
T= 2pi time square root of (l /g)
frequency= 1/t
frequency= 1/ [ 2pi time square root of (l /g) ]

I need comment about my solution. Thanks

 

[learningphysics]

A ball is attached to a string with length of L. It swings in a horizontal circle, with a constant speed. The string makes an angle (theta) with the vertical, and T is the magnitude of the tension in the string.

1)Determine the Mass of the Ball.
2)Determine th Speed of the Ball.
3)Determine the Frequency of revolutions of the Ball.

1)
F=Tsin(theta)=ma
m= (Tcos theta)/g

Looks good.

2)
F= (mv^2)/a= Tsin(theta)

don't you mean F = mv^2/r =Tsin(theta) ? and what is r?

also, your final answer should not have m in it... you should write your answer in terms of the given values of L, theta and T.

3)
T= 2pi time square root of (l /g)
frequency= 1/t
frequency= 1/ [ 2pi time square root of (l /g) ]

I need comment about my solution. Thanks

Not sure about part 3 (might be right)... I think it's better to use your v value to derive the frequency. try to use the v value along with the circumference of the circle to get the frequency... might come out to what you have above, in which case you've double checked your answer.

 

[logglypop]

I got new approach on number 2

Tcos(theta)- mg= 0
T= mg/[cos(theta)]

F=Tsin(theta)
F= mg/[cos(theta)] * sin(theta) subtitute for T
F=mgtan(theta)

F=(mv^2)/r r=Lsin(theta)
mgtan(theta)=(mv^2)/ Lsin(theta)
v= square root of gLtan(theta)sin(theta)

 

[logglypop]

please give me some advice

 

[learningphysics]

I got new approach on number 2

Tcos(theta)- mg= 0
T= mg/[cos(theta)]

F=Tsin(theta)
F= mg/[cos(theta)] * sin(theta) subtitute for T
F=mgtan(theta)

F=(mv^2)/r r=Lsin(theta)
mgtan(theta)=(mv^2)/ Lsin(theta)
v= square root of gLtan(theta)sin(theta)

Yes, that looks right to me. now use this v value to get frequency in part 3.

period = 2*pi*r/v