A solid cylinder of mass m and radius R rolls down a parabolic path PQR from height without slipping (assume R ≪ H) as shown in the figure below. Path PQ is rough (and so the cylinder will roll on that path), whereas path QR is smooth, or frictionless (so the cylinder will only slide, not roll, in this region). Determine the height h above point Q reached by the cylinder on path QR. (Use the following as necessary: m, g, H, and R.)
figure dicribtion: point P is at the highest point on the left of the parabolic path . Point Q is at the lowest piont of the
PE going into the down side: mgH
KE is translational, and rotational or
mgH=1/2 m v^2 + 1/2 Imr^2*w^2
mgH=1/2 m v^2 + 1/2 (1/2)mr^2*w^2
but when rolling, w=v/r
mgH=1/2 m v^2+1/4*mv^2 =3/4mv^2
On the second side, rolling doesn t happen, the shell just spins, and how high it goes depends on the initial translational energy (1/2 m v^2= 1/2mgH*4/3)=2/3mgH. 2/3mgH=mgR . R=2/3H
Figure description: point P is at the highest point on the left of the parabolic path . Point Q is at the lowest point of the parabolic path and point R is at the highest point on the right of the parabolic path. The height is from the ground to point P.
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PE going into the down side: mgH
KE is translational, and rotational or
mgH=1/2 m v^2 + 1/2 Imr^2*w^2
mgH=1/2 m v^2 + 1/2 (1/2)mr^2*w^2
but when rolling, w=v/r
mgH=1/2 m v^2+1/4*mv^2
On the second side, rolling doesn t happen, the shell just spins, and how high it goes depends on the initial translational energy (1/2 m v^2=1/4 m gH*4/1)=2mgH. 2mgH=mgR . R=2H
PE going into the down side: mgH
KE is translational, and rotational or
mgH=1/2 m v^2 + 1/2 Imr^2*w^2
mgH=1/2 m v^2 + 1/2 (1/2)mr^2*w^2
but when rolling, w=v/r
mgH=1/2 m v^2+1/4*mv^2 =3/4mv^2
On the second side, rolling doesn t happen, the shell just spins, and how high it goes depends on the initial translational energy (1/2 m v^2= 1/2mgH*4/3)=.6mgH. .6mgH=mgR . R=.6H
To determine the height h above point Q reached by the cylinder on path QR, we can use the principles of conservation of mechanical energy.
1. Calculate the potential energy at point P:
The potential energy at point P is given by mgh, where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the total height of the parabolic path.
2. Calculate the kinetic energy at point P:
Since the cylinder is rolling without slipping, the kinetic energy at point P is given by (1/2)Iω^2, where I is the moment of inertia of the cylinder and ω is its angular velocity.
The moment of inertia of a solid cylinder about its axis of rotation is given by (1/2)mR^2.
The angular velocity ω can be calculated using the formula ω = v/R, where v is the linear velocity of the cylinder at point P. Since the cylinder is rolling without slipping, v = Rω.
Substituting the value of v in the formula for kinetic energy, we get (1/2)(1/2)mR^2(Rω/R)^2 = (1/4)mR^2ω^2.
3. Calculate the total energy at point P:
The total energy at point P is the sum of the potential energy and kinetic energy, i.e., mgh + (1/4)mR^2ω^2.
4. Calculate the kinetic energy at point Q:
Since the path QR is smooth and frictionless, the cylinder only slides without rolling. Therefore, at point Q, the kinetic energy is given by (1/2)mv^2, where v is the linear velocity of the cylinder at point Q.
Using the principle of conservation of mechanical energy, the total energy at point P must be equal to the kinetic energy at point Q.
Setting mgh + (1/4)mR^2ω^2 = (1/2)mv^2, we can solve for v.
5. Calculate the height h above point Q:
Once we have calculated the linear velocity v at point Q, we can use the conservation of mechanical energy again to find the height h above point Q.
The kinetic energy at point Q is given by (1/2)mv^2, and since the potential energy is zero at point Q, the total energy is equal to the kinetic energy.
Therefore, (1/2)mv^2 = (1/2)mv^2 + (1/4)mR^2ω^2.
Solving for h, we find h = (v^2 - R^2ω^2)/(2g).
This equation gives us the height h above point Q reached by the cylinder on path QR.
Note: The assumption R ≪ H is given in the problem statement, and it implies that the radius of the cylinder is much less than the total height of the parabolic path.
Am I not understanding something here? on the way up, it is smooth, but does not roll. It should continue spinning with the same rotational KE it had at Q, so the final PE at the height h will be determined by the translational KE at the bottom.
So figure the fall PE mgh divides up into translational KE and rotational KE, and they are related because it is rolling...I assume you know how to relate w to speed. then at the bottom, it changes, only the translational KE turns into PE of height P