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(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!
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
To find the maximum height the object will reach after leaving the ramp, we can follow these steps:
1. Start by determining the potential energy at the height H. The potential energy (PE) is given by PE = m • g • H, where m is the mass of the object, g is the acceleration due to gravity, and H is the initial height. In this case, H = 9.0 m.
2. Next, calculate the total energy at height R. The total energy (E) is the sum of the potential energy, the rotational kinetic energy, and the translational kinetic energy. In this case, the rotational kinetic energy is given by (1/2) • I • ω^2, where I is the moment of inertia of the object and ω is its angular velocity. The translational kinetic energy is given by (1/2) • m • v^2, where v is the linear velocity of the object.
3. We know that the moment of inertia of the object is given by I = c • m • r^2, where c is a constant and r is the radius of the object. In this case, we are given that c = 0.39.
4. Using the law of conservation of energy (PE = E), we can set up the equation m • g • H = m • g • R + (1/2) • (c • m • r^2) • ω^2 + (1/2) • m • v^2.
5. Solve this equation for v by substituting the given values: H = 9.0 m, R = 3.0 m, g = 9.8 m/s^2. This will give you an equation in terms of v.
6. Use the equation v = sqrt( g(H-R) / (1.39 • 0.5) ) to calculate the value of v. Substitute the values of g, H, and R into this equation to find v.
7. Finally, use the equation h = (v^2) / (2 • g) to calculate the maximum height the object will reach after leaving the ramp. Substitute the values of v and g into this equation to find h.
By following these steps and performing the necessary calculations, you should be able to find the correct answer.