I don't know why my answer is wrong for this question. My answer is .8H. If you can help me,please do so because I got stuck at this question for a whole week. Please show me with steps. If you got ignore by the same question, I'm more ignore by it then you.
A solid cylinder 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 friction-less (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 solid cylinder 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
bobpursley Monday, April 16, 2018 at
This is how I have tried to solve it base on what I had learned from a previous post.
mgh=1/2mv^2+1/2(1/4mr^2)*w^2
mgh==1/2mv^2+1/8mv^2
v^2=(gh)/(5/8)
(1/2 m v^2=1/2 m gH*8/5)=.8H
https://www.jiskha.com/display.cgi?id=1523916313
Is it possible you can put your name in the post, so we can track who asked what?
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
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
thank you
To solve this problem, let's break it down step by step.
Step 1: Determine the initial potential energy
The initial potential energy of the cylinder at point P can be calculated as mgh, where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the height from the floor to point P.
Step 2: Determine the final kinetic energy
The final kinetic energy of the cylinder at point Q can be calculated as 1/2mv^2, where v is the velocity of the cylinder at point Q.
Step 3: Determine the rotational kinetic energy
Since the cylinder is rolling without slipping on path PQ, it has rotational kinetic energy as well. The rotational kinetic energy of the cylinder can be calculated as 1/2(1/4mr^2)*ω^2, where r is the radius of the cylinder and ω is the angular velocity of the cylinder.
Step 4: Equate the initial potential energy to the sum of final kinetic energy and rotational kinetic energy.
Setting up an equation by equating the initial potential energy to the sum of final kinetic energy and rotational kinetic energy, we have:
mgh = 1/2mv^2 + 1/2(1/4mr^2)*ω^2
Step 5: Simplify the equation
Since we are interested in the value of h above point Q reached by the cylinder on path QR, we can ignore the terms related to rotation because path QR is smooth and frictionless. This means there is no rotation on path QR. Therefore, the equation simplifies to:
mgh = 1/2mv^2
Step 6: Solve for h
To solve for h, isolate it by dividing both sides of the equation by mg:
h = v^2 / (2g)
Step 7: Substitute the value of v
Using the relationship between linear velocity v and angular velocity ω for a rolling cylinder, we have v = ωr. Substituting this into the equation, we get:
h = (ωr)^2 / (2g)
Step 8: Substitute the value of ω
Since path QR is smooth, the cylinder is not rolling on this path and only slides. This means ω = 0 for path QR. Substituting this into the equation, we get:
h = (0)^2 / (2g)
h = 0
Thus, the height h above point Q reached by the cylinder on path QR is 0.