Question
Only need the answer for part E, please help! :-(
v2 = 21.7
= 98.6
(25%) Problem 1: A thin cylindrical ring starts from rest at a height 21-86 m. The ring has a radius R 22 cm and a mass M= 3 kg R: 2 ©theexpertta.com 20% Part (a) Write an expression for the ring's initial energy at point 1, assuming that the gravitational potential energy at point 3 s zero 20% Part (b) If the ring rolls (without slipping) all the way to point 2, what is the ring's energy at point 2 in terms of h2 and v2? 20% Part (c) Given h2 = 38 m, what is the velocity of the ring at point 2 in ms? 20% Part (d) What is the ring's rotational velocity in rad/s? × 20% Part (e) After passing point 2 the hill becomes frictionless and the ring's rotational velocity remains constant. What is the linear velocity of the ring at point 3 in m/s? Grade Summary Deductions Potential 15% 8596 sin0 cotan0asin acos0 atan0 acotan sinh0 cosh0 tanh cot co tan Submissions Attempts remaining:;4 (5% per attempt) detailed view END 5% 5% e Degrees Radians PACE
Explanation / Answer
from 1 to 2:
PEi + KEi = PEf + KEf
m g h1 + 0 = m g h2 + (m v2^2 /2 + I w^2 /2 )
m g (h1 - h2) = m v^2 /2 + (m r^2)(v/r)^2 /2
v2^2 = g (h1 - h2) = 48 g
now from 1 to 3:
m g h1+ 0 = 0 + (m v3^2 / 2) + (rot KE )
m g h1 = m v3^2 /2 + (m v2^2 /2)
m g (86) = m v3^2 /2 + m (48 g) / 2
86 g = v3^3/2 + 24g
v43 = sqrt(2 x9.8 x (86 - 24) ) = 34.9 m/s ........Ans
