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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
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(H-R)/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(9-3)/1.39x0.5)
Sorry for reposting this, but i need immediate response please!

  • Physics - ,

    v =sqrt( g(H-R)/(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

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