A child is twirling a 0.0116 kg ball on a string in a horizontal circle whose radius is 0.102 m. The ball travels once around the circle in 0.560 s.
(a) Determine the centripetal force acting on the ball.
N
(b) If the speed is doubled, does the centripetal force double?
yes
no
If not, by what factor does the centripetal force increase?
I was able to work the problem but i'm not sure what it means by factor for the third part of the problem. Please help.
Centripetal force is
M V^2/R
In your case V = 2 pi R /(0.56 s) = 1.1144 m/s, so
Force = (0.0116)(1.1144)^2/(0.102)= ? N
If V is charged to 2V, the Force becomes 4 times as large, because the 2 gets squared. The factor they are looking for is 4.
To find the centripetal force acting on the ball, you can use the formula:
Fc = (m * v^2) / r
Where:
- Fc is the centripetal force
- m is the mass of the ball (0.0116 kg)
- v is the velocity of the ball, which can be found using the formula v = 2πr / T, where T is the time taken to complete one revolution around the circle (0.560 s)
- r is the radius of the circle (0.102 m)
(a) Plug the given values into the formula to calculate the centripetal force:
Fc = (0.0116 kg * ((2π * 0.102 m) / 0.560 s)^2) / 0.102 m
Solve this equation to find the value of the centripetal force in Newtons (N).
(b) If the speed is doubled, the centripetal force does not double. The centripetal force depends on the square of the velocity, so if the speed is doubled, the centripetal force will be four times greater.
To find the factor by which the centripetal force increases, divide the new centripetal force by the original centripetal force. In this case, it would be:
Factor = New centripetal force / Original centripetal force
Substitute the new velocity into the formula and calculate the new centripetal force. Then divide the new centripetal force by the original centripetal force to find the factor by which it increases.