- Questions:1. A uniform disk is rolled down a hill of height 11m. What is the speed of the disk when it reaches the bottom of the hill?
2. As soon as the disk reaches the bottom of the hill, it hits a horizontal surface of frictional coefficient 0.13. What is the acceleration of the disk?
3. A ball of mass 2kg is attached to a string of length 0.75m and rotated vertically with an angular velocity of 7rad/s. What is the ratio of the string tension at the top of the loop to the string tension at the bottom?
Ignore significant figures.
- Answer 1) g * (11 m) = 0.5v^2
2) g * 0.13 = 1.275 m/s^2
1.275 m/s^2 against the motion of the disk
3)F + (2 kg) * g = (7 rad/s * (0.75 m/rad))^2 / (0.75 m)
F = 17.13 N
F - (2 kg) * g = (7 rad/s * (0.75 m/rad))^2 / (0.75 m)
F = 56.37 N