Physics
posted by Adam .
A small circular object with mass m and radius r has a moment of inertia given by
I = cmr2.
The object rolls without slipping along the track shown in the figure. The track ends with a ramp of height R = 3.0 m that launches the object vertically. The object starts from a height H = 9.0 m. To what maximum height will it rise after leaving the ramp if c = 0.39?
I am really confused about this, please help thanks!
I really need someone please help; thanks!
Physics please help  Elena, Thursday, March 29, 2012 at 6:07pm
Moment of inertia of the circular object is I =cmr^2
c=0.39
The potential energy at the height H is PE=m•g•H
The total energy at height R is
E=m•g•R+(1/2) •I•ω^2 +( 1/2) •m•v^2 =
=m•g•R + (1/2) • (0.39) •m•r^2 ω^2+(1/2) •m•v^2 =
=m•g•R + (1/2) • (0.39) •m•r^2 v^2/r^2+(1/2) •m•v^2 =
=m•g•R + (1/2) •m•v^2 •1.39
According to the law of conservation of energy PE =E :
m•g•H = m•g•R+(1.39)•(1/2)•m•v^2
Now H =9 m, R=3.0 m, g = 9.8m/s^2,
v =sqrt( g(HR)/1.39•0.5)
Then maximum height the object moves after leaving the track is
h = (v^2)/2g
Substitute the value of g and v in the above equation and solve for h
Physics please help  Adam, Thursday, March 29, 2012 at 7:48pm
I keep getting 1.079 m, but that's wrong I don't know what I am doing wrong I did as you said h=(4.599)^2/2x9.8 but its wrong! And I got 4.599 from v=sqrt(9.8(93)/1.39x0.5)
Sorry for reposting this, but i need immediate response please!

v =sqrt( g(HR)/(1.39•0.5)) = sqrt((9.8•6)/(1.39•0.5))=9.2 m/s,
h = (v^2)/2g =(9.2)^2/(2•9.8) =4.32 m
Respond to this Question
Similar Questions

PHYSICS
A spherically symmetric object with radius of .7m and mass of 1.6kg rolls without slipping accross a horizontal surface with velocity of 1.7m/s. It then rolls up an invline with an angle of 28degrees and comes to rest a distance d … 
physics
A small circular object with mass m and radius r has a moment of inertia given by I = cmr2. The object rolls without slipping along the track shown in the figure. The track ends with a ramp of height R = 3.0 m that launches the object … 
Physics
A small circular object with mass m and radius r has a moment of inertia given by I = cmr2. The object rolls without slipping along the track shown in the figure. The track ends with a ramp of height R = 3.0 m that launches the object … 
Physics please help
A small circular object with mass m and radius r has a moment of inertia given by I = cmr2. The object rolls without slipping along the track shown in the figure. The track ends with a ramp of height R = 3.0 m that launches the object … 
Physics
A small circular object with mass m and radius r has a moment of inertia given by I = cmr^2. The object rolls without slipping along the track shown in the figure. The track ends with a ramp of height R = 3.0 m that launches the object … 
Physics
A sphere with a moment of inertia 0.658mr2 and mass m and radius r rolls without slipping along the track shown on a planet where the acceleration due to gravity is g. It starts from rest at a height h = 4.87R above the bottom of the … 
Physics
What is the moment of inertia of an object that rolls without slipping down a 2.00mhigh incline starting from rest, and has a final velocity of 6.00 m/s? 
MITx: 8.01x Classical Mechanics
An object of mass m is released from rest at a height h above the surface of a table. The object slides along the inside of the looptheloop track consisting of a ramp and a circular loop of radius R shown in the figure. Assume that … 
Physics Rolling Motion!!
A spherically symmetric object with radius R= 0.51m and mass of M= 3.5kg rolls without slipping accross a horizontal surface with velocity of V = 4.4m/s. It then rolls up an incline with an angle of 30 degrees and comes to rest a distance … 
Physics help Rolling Motion!!
A spherically symmetric object with radius R= 0.51m and mass of M= 3.5kg rolls without slipping accross a horizontal surface with velocity of V = 4.4m/s. It then rolls up an incline with an angle of 30 degrees and comes to rest a distance …