- #1
PEToronto
- 7
- 0
Homework Statement
A 200g ball on a 55-cm-long string is swung in a vertical circle about a point 200 cm above the floor. The string suddenly breaks when it is parallel to the ground and the ball is moving upward. The ball reaches a height 600 cm above the floor. What was the tension in the string an instant before it broke?
Given variables:
initial height: 0.2 m
final height: 0.6 m
radius: 0.55 m
mass of ball: 0.2 kg
gravity: -9.8 m/s^2
The attempt at a solution
My answer: The ball is rotating around a point that is 200 cm (0.2 m) above the ground, and snaps off when the string is parallel to the ground- i.e. when the ball is the same height as the center-point of its rotation. It reaches a height of 0.6 m or 600 cm, with a total vertical displacement of (0.6-0.2) m, or 0.4 m, upward. During this displacement of 0.4 m, it has a negative acceleration of 9.8 m/s^2 due to gravity. Using this information, and assuming that the velocity at maximum height is equal to zero, I can use the following equation to figure out the tangential velocity at the instant the ball broke off the string:
Vf^2 = Vi^2 + 2a(y2-y1)
=> 0^2 = Vi^2 + 2(-9.8)(0.6-0.2)
=> 0 = Vi^2 + 2(-9.8)(0.4)
=> -Vi^2 = -7.84
=> Vi = sqrt(7.84) = 2.8
Now I know that the tangential velocity of the ball the instant before it broke off the string was 2.8 m/s upward.
Since centripetal acceleration is "velocity squared over radius", and radius is 55 cm or 0.55 m, so
ac = v^2 / r
=> ac = (2.8)^2 / (0.55) = 14.25
With an acceleration of 14.25 m/s^2 toward the center, the force toward the center should be mass times acceleration, and the mass of the ball is 0.2 kg so
Fc = ac * m
=> 14.25 * 0.2 = 2.85
So the tension on the string should be 2.85 N.
Where did I go wrong?
Did I neglect some vertical component of tension that counteracts gravity while the ball is still attached to the string?
A 200g ball on a 55-cm-long string is swung in a vertical circle about a point 200 cm above the floor. The string suddenly breaks when it is parallel to the ground and the ball is moving upward. The ball reaches a height 600 cm above the floor. What was the tension in the string an instant before it broke?
Given variables:
initial height: 0.2 m
final height: 0.6 m
radius: 0.55 m
mass of ball: 0.2 kg
gravity: -9.8 m/s^2
The attempt at a solution
My answer: The ball is rotating around a point that is 200 cm (0.2 m) above the ground, and snaps off when the string is parallel to the ground- i.e. when the ball is the same height as the center-point of its rotation. It reaches a height of 0.6 m or 600 cm, with a total vertical displacement of (0.6-0.2) m, or 0.4 m, upward. During this displacement of 0.4 m, it has a negative acceleration of 9.8 m/s^2 due to gravity. Using this information, and assuming that the velocity at maximum height is equal to zero, I can use the following equation to figure out the tangential velocity at the instant the ball broke off the string:
Vf^2 = Vi^2 + 2a(y2-y1)
=> 0^2 = Vi^2 + 2(-9.8)(0.6-0.2)
=> 0 = Vi^2 + 2(-9.8)(0.4)
=> -Vi^2 = -7.84
=> Vi = sqrt(7.84) = 2.8
Now I know that the tangential velocity of the ball the instant before it broke off the string was 2.8 m/s upward.
Since centripetal acceleration is "velocity squared over radius", and radius is 55 cm or 0.55 m, so
ac = v^2 / r
=> ac = (2.8)^2 / (0.55) = 14.25
With an acceleration of 14.25 m/s^2 toward the center, the force toward the center should be mass times acceleration, and the mass of the ball is 0.2 kg so
Fc = ac * m
=> 14.25 * 0.2 = 2.85
So the tension on the string should be 2.85 N.
Where did I go wrong?
Did I neglect some vertical component of tension that counteracts gravity while the ball is still attached to the string?
