Can somebody please shows me how to get to the answer key,06H?
A thin, spherical shell of mass m and radius R rolls down a parabolic path PQR from height H 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.)
The spherical shell starts on the left side of the parabola curve at point P then goes down to point Q then up to point R.H is from the floor to point P
PE going into the down side: mgH
KE is translational, and rotational or
mgH=1/2 m v^2 + 1/2 (2/3)mr^2*w^2
but when rolling, w=v/r
mgH=1/2 m v^2+1/3*mv^2
v^2=gh/(5/6) check me on that.
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=1/2 m gH*6/5)=.6mgH
so that inergy will equal how high it goes.
.6mgH=mgR
R=.6H I have no idea where the 06H came from. Now a question for you. I outlined all this on your post a couple of days ago.
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 or .7H
To determine the height h above point Q reached by the cylinder on path QR, you can follow these steps:
Step 1: Analyze the initial and final states of the spherical shell on path QR. At point Q, the shell has a kinetic energy and potential energy. At point R, the shell only has potential energy.
Step 2: Apply the conservation of mechanical energy principle. Since there is no friction on path QR, mechanical energy is conserved. The initial mechanical energy at point Q is the sum of its kinetic energy (1/2mv^2) and potential energy (mgH), where v is the velocity of the shell at point Q and g is the acceleration due to gravity.
Step 3: At point R, the only remaining energy is potential energy. Since the shape of the path is a parabola, the final potential energy at height h above point Q is mg(h+H).
Step 4: Equate the initial and final mechanical energies (kinetic + potential = potential). This gives you the equation:
1/2mv^2 + mgh = mgh + mgh + mgH
Step 5: Simplify the equation by canceling the common terms:
1/2v^2 = 2gH
Step 6: Solve for v^2:
v^2 = 4gH
Step 7: Take the square root to find the velocity of the shell at point Q:
v = 2√(gH)
Step 8: Apply the conservation of linear momentum principle. Since there is no net external force acting on the shell, linear momentum is conserved. The initial linear momentum at point Q is the product of mass and velocity, which is mv.
Step 9: At point R, the linear momentum is zero since the shell comes to rest and has no velocity.
Step 10: Equate the initial and final linear momenta to find m:
mv = 0
Step 11: Solve for m:
m = 0 (since m ≠ 0)
Step 12: Now that we have determined m = 0, we can substitute this value into the equation for potential energy at point R:
Potential energy at R = mg(h+H)
= 0 (since m = 0)
Therefore, the height h above point Q reached by the cylinder on path QR is h = 0.