JEE Challenger
More from System of Particles and Rotational Motion
PhyHardAdv 2022 • Paper 2

Angular Momentum of Rolling Disk on Circular Path

Consider a flat surface of a thin uniform disk AA having radius RR that is fixed to a horizontal table. Another thin uniform disk BB with mass MM and radius RR rolls without slipping along the circumference of AA, as illustrated in the figure. A flat surface of BB also lies on the plane of the table. The center of mass of disk BB rotates with a constant angular speed ω\omega around the vertical axis passing through the center of AA. If the angular momentum of BB relative to the center of AA is given by nMωR2nM\omega R^2, then what is the value of nn?

Question Diagram 1

Options

A

2

B

5

Correct
C

72\frac{7}{2}

D

92\frac{9}{2}

Topics & Concepts

System of Particles and Rotational MotionMoment of InertiaAngular MomentumRolling Motion

Step-by-Step Solution

To find the value of nn, we calculate the angular momentum of disk BB relative to the center of fixed disk AA (denoted as OAO_A).

1. Kinematics of Disk BB

  • Radius of disk A=RA = R
  • Radius of disk B=RB = R
  • Distance from the center of disk AA (OAO_A) to the center of mass of disk BB (OBO_B) is: rcm=R+R=2Rr_{\text{cm}} = R + R = 2R

The center of mass OBO_B rotates about OAO_A with a constant angular speed ω\omega. Therefore, the magnitude of the linear velocity of OBO_B is: vcm=rcmω=2Rωv_{\text{cm}} = r_{\text{cm}} \omega = 2R\omega

In vector form, using polar coordinates centered at OAO_A: r⃗cm=2Rr^\vec{r}_{\text{cm}} = 2R \hat{r} v⃗cm=2Rωθ^\vec{v}_{\text{cm}} = 2R\omega \hat{\theta}

2. Angular Velocity of Disk BB (ωB\omega_B)

Let ω⃗B=ωBk^\vec{\omega}_B = \omega_B \hat{k} be the absolute spin angular velocity of disk BB about its own center of mass OBO_B.

The point of contact PP on disk BB is located at a vector position r⃗P/OB=−Rr^\vec{r}_{P/O_B} = -R \hat{r} relative to OBO_B. Since disk BB rolls without slipping on the fixed disk AA, the velocity of point PP must be zero: v⃗P=v⃗cm+ω⃗B×r⃗P/OB=0⃗\vec{v}_P = \vec{v}_{\text{cm}} + \vec{\omega}_B \times \vec{r}_{P/O_B} = \vec{0}

Substituting the known values: 2Rωθ^+(ωBk^)×(−Rr^)=0⃗2R\omega \hat{\theta} + (\omega_B \hat{k}) \times (-R \hat{r}) = \vec{0} 2Rωθ^−ωBRθ^=0⃗2R\omega \hat{\theta} - \omega_B R \hat{\theta} = \vec{0}

Solving for ωB\omega_B: ωB=2ω\omega_B = 2\omega

Thus, the absolute angular velocity of disk BB is ω⃗B=2ωk^\vec{\omega}_B = 2\omega \hat{k}.

3. Angular Momentum of Disk BB Relative to OAO_A

By Koenig's theorem, the total angular momentum L⃗\vec{L} of a rigid body about an origin OAO_A is the sum of its spin angular momentum about its center of mass and the orbital angular momentum of its center of mass relative to OAO_A: L⃗=L⃗cm+r⃗cm×p⃗cm\vec{L} = \vec{L}_{\text{cm}} + \vec{r}_{\text{cm}} \times \vec{p}_{\text{cm}}

  1. Spin Angular Momentum (L⃗cm\vec{L}_{\text{cm}}): For a thin uniform disk of mass MM and radius RR, the moment of inertia about its central vertical axis is Icm=12MR2I_{\text{cm}} = \frac{1}{2} M R^2. L⃗cm=Icmω⃗B=(12MR2)(2ωk^)=MR2ωk^\vec{L}_{\text{cm}} = I_{\text{cm}} \vec{\omega}_B = \left(\frac{1}{2} M R^2\right) (2\omega \hat{k}) = M R^2 \omega \hat{k}

  2. Orbital Angular Momentum (r⃗cm×p⃗cm\vec{r}_{\text{cm}} \times \vec{p}_{\text{cm}}): r⃗cm×p⃗cm=(2Rr^)×(M⋅2Rωθ^)=4MR2ωk^\vec{r}_{\text{cm}} \times \vec{p}_{\text{cm}} = (2R \hat{r}) \times (M \cdot 2R\omega \hat{\theta}) = 4 M R^2 \omega \hat{k}

Combining both parts: L⃗=MR2ωk^+4MR2ωk^=5MR2ωk^\vec{L} = M R^2 \omega \hat{k} + 4 M R^2 \omega \hat{k} = 5 M R^2 \omega \hat{k}

The magnitude of the angular momentum of disk BB relative to the center of AA is: L=5MωR2L = 5 M \omega R^2

Comparing this with nMωR2nM\omega R^2, we find: n=5n = 5

Correct Answer: (B) 5

PreviousNext
Angular Momentum of Rolling Disk on Circular Path | Physics PYQ Solution - JEE Challenger