Ask questions and get helpful answers.

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

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩

5 answers

  1. 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.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  2. 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

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  3. ?
    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

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  4. 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

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  5. 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

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Answer this Question

Related Questions

Still need help?

You can ask a new question or browse existing questions.