Class Management Help Homework #4 Begin Date: 1/15/2018 12:00:00 AM-Due Date: 2/
ID: 1871887 • Letter: C
Question
Class Management Help Homework #4 Begin Date: 1/15/2018 12:00:00 AM-Due Date: 2/18/2018 11:59:00 PM End Date: 2/25/2018 12:00:00 AM (596) Problem 8: Two blocks, which can be modeled as point masses, are connected by a massless string which passes through a hole in a frictionless table. A tube extends out of the hole in the table so that the portion of the string between the hole and MI remains parallel to the top of the table. The blocks have masses M1 = 1.7 kg and M2 = 2.2 kg. Block l is a distance = 0.65 m from the center of the frictionless surface. Block 2 hangs vertically underneath Otheexpertta.com X 50% Part (a) Assume that block two. M. does not move relative to the table and that block one. M1, is rotating around the table. what is the speed of block one, Mi, in meters per second? Grade Summary Deductions Potential 1-1 .42 5% 95% sin() cosO 789 HOME Submissions Attempts remaining: 4 (5% per attempt) detailed view cotan as acos) atanacotan sinh0 cosh)th cotanh 0 END 5% O Deg rees Radians BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 2% deduction per hint. Hints remaining: 2 Feedback: 2% deduction per feedback. 50% Part (b) How much time, in seconds, does it take for block one, MI, to make one revolution?Explanation / Answer
a) To bring the whole system in equilibrium, the weight of mass M2 must provide centripetal acceleration to M1.
M2*g = M1v2 / r
solve for v, we get v2 = M2*g*r / M1
Putting the given values,
v2 = 2.2*9.8*0.65 / 1.7 = 8.2435
v = 2.8712 m/s
b) time = distance / velocity
time = 2*pi*r / v
time = 4.08407 / 2.8712
time = 1.422 seconds.