- #1
warmfire540
- 53
- 0
A 2.0 kg ball is attached to a vertical post with two strings, one 2.0 m long and the other 1.0 m long as shown in the figure. If the ball is set whirling in a horizontal circle, what is the minimum speed necessary for the lower string to be taught? If the ball has a constant speed of 6 ms-1, find the tension on each string.
|\
| \
| \ -2.0m
| \
| \
|----0 -2.0kg
|
| ^1m
For the first part, figured that the 2m string has two components of tension Tsin60 and Tcos60.
Tsin60=mg=19.6
Tcos60=mv^2/r=2v^2
tan60=19.6/2v^2
v=5.66 m/s
is that right? I calculated the angle that the 2m string made w/ the x-axis and went from there...
Now..for the second part..
i figured that the tension in the 1m string is the centripital force which is:
T=mv^2/r=2*36/1= 72N
Now for the second string..what would the tension be?
i'm guessing i have to use the Tsin60 and the Tcos60
Tsin60=19.6N
Tcos60=72N
T=sqrt(19.6^2+72^2)
T=74.6N
I hope this is right, i did all the work, I'm just looking for confirmation please
|\
| \
| \ -2.0m
| \
| \
|----0 -2.0kg
|
| ^1m
For the first part, figured that the 2m string has two components of tension Tsin60 and Tcos60.
Tsin60=mg=19.6
Tcos60=mv^2/r=2v^2
tan60=19.6/2v^2
v=5.66 m/s
is that right? I calculated the angle that the 2m string made w/ the x-axis and went from there...
Now..for the second part..
i figured that the tension in the 1m string is the centripital force which is:
T=mv^2/r=2*36/1= 72N
Now for the second string..what would the tension be?
i'm guessing i have to use the Tsin60 and the Tcos60
Tsin60=19.6N
Tcos60=72N
T=sqrt(19.6^2+72^2)
T=74.6N
I hope this is right, i did all the work, I'm just looking for confirmation please