Question

A cylinder of mass 0.3 kg and radius 8 cm (and I = 1/2 mR2 ) is placed at the top of a long, straight ramp of length 2 meters. The cylinder is released from rest and allowed to roll without slipping to the bottom. When it reaches the bottom, it is rotating at a rate of 6.5 rev/s. Answers included. Need the process.

a) What is the linear speed at the bottom of the ramp? 3.3m/s

b) What is the total KE of the cylinder at the bottom of the ramp? 2.4J

c) Use conservation of mechanical energy for the Earth-cylinder system to find the height of the ramp. .8m

d) What were the linear and angular accelerations of the cylinder down the incline (hint: use kinematics)? 2.7m/s^2

Answer 1

mass of cylindr = m = 0.3Kg; Radius of cylinder = R = 0.08mrs

Ramp length = 2meters

Intial velocity of cylinder when released = 0 m/s ,angular velocity at the bottom of ramp = 6.5 rev/s

a) Liner speed of the Cylinder at bottom of ramp = angular speed in Rad/s X R

= anglar speed in rev/s X 2 pi X R

= 6.5 Rev/s X 2 X pi X 0.08 m =3.3 m/s

b) KE of cylinder at bottom of ramp =

= 0.5 X (0.5 mR^2) ()^2 + 0.5 X m X

=(1/4) mVcm^2 + (1/2) Vcm^2 = (3/4) m Vcm^2

= 2.4 J

c) Initially the cylinder has only PE = mgh

Finally at bottom of ramp this PE is totally converted to KE = (3/4) m Vcm^2 =

mgh = (3/4) m Vcm^2

h = 0.75 X Vcm^2/g = 0.8 m

d) Let us consider that as the cylinder rolls down it has weight mg acting vertically downwards. The normal reaction offered by inclined plane is N and is angle of the ramp with horizontal and f is friction force on cylinder due to plane at point of contact

N- mg cos =0 , for motion normal to the incline

mgsin -f = ma , for motion along the incline

Now for rotational motion

Torque =

fR =

Since and

a= (2/3) g sin ( from2) = (2/3) 9.8 m/s^2 X (0.8/2) =2.6 m/s^2

= 2.6 /0.08 = 32rad/s^2