Question

A disc of circumference 16 cm stands vertically on horizontal surface. A horizontal force P acts on centre of disc. Half of circumference ABC is rough and friction is sufficient to prevent slipping when disc rolls along xy. The other half of circumference (ADC) is smooth. The disc starts from rest and point C is at bottom. Find the distance (in cm) moved by center of disc along xy when if completes one rotation.

Answer 1

19

Answer 2

The problem describes a disc rolling on a horizontal surface under the action of a horizontal force P at its center. The circumference of the disc is 16 cm. Half of the circumference (ABC) is rough, allowing for rolling without slipping, while the other half (ADC) is smooth. The disc starts from rest with point C at the bottom and completes one full rotation. We need to find the total distance moved by the center of the disc.

First, let's determine the radius of the disc: Circumference $C = 2\pi R = 16$ cm. So, the radius $R = \frac{16}{2\pi} = \frac{8}{\pi}$ cm.

The motion can be divided into two phases, each covering half a rotation ($\pi$ radians):

Phase 1: Rough surface (ABC) in contact with the ground (from C to A at bottom)

• The disc starts from rest.

• The surface is rough, and friction is sufficient to prevent slipping. This means the disc rolls without slipping.

• When a disc rolls without slipping, the distance moved by its center of mass is equal to the arc length of the circumference that has come into contact with the ground.

• Since this phase covers half a rotation, the angular displacement is $\Delta\theta_1 = \pi$ radians.

• The distance moved by the center of mass in this phase, $x_1$, is given by: $x_1 = R \Delta\theta_1 = R\pi = \left(\frac{8}{\pi}\right)\pi = 8$ cm.

• Let's analyze the dynamics in this phase to find the velocity at the end of this phase.

  • Linear equation of motion: $P - f = Ma_1$ (where $f$ is the static friction force, opposing the tendency to slip forward due to P).
  • Rotational equation of motion about the center of mass: $fR = I\alpha_1$. For a disc, $I = \frac{1}{2}MR^2$. So, $fR = \frac{1}{2}MR^2\alpha_1 \implies f = \frac{1}{2}MR\alpha_1$.
  • No-slip condition: $a_1 = R\alpha_1$. So, $f = \frac{1}{2}Ma_1$.
  • Substitute $f$ into the linear equation: $P - \frac{1}{2}Ma_1 = Ma_1 \implies P = \frac{3}{2}Ma_1$.
  • Therefore, the acceleration of the center of mass in this phase is $a_1 = \frac{2P}{3M}$.

• The velocity of the center of mass at the end of Phase 1 (when point A is at the bottom), $v_1$, can be found using kinematics: $v_1^2 = u^2 + 2a_1 x_1$. Since $u=0$: $v_1^2 = 2a_1 x_1 = 2\left(\frac{2P}{3M}\right)(8) = \frac{32P}{3M}$.

• The angular velocity at the end of Phase 1 is $\omega_1 = \frac{v_1}{R}$.

Phase 2: Smooth surface (ADC) in contact with the ground (from A to C at bottom)

• This phase also covers half a rotation, so the angular displacement is $\Delta\theta_2 = \pi$ radians.
• The surface is smooth, meaning there is no friction ($f=0$).
• Linear equation of motion: $P = Ma_2$. So, the acceleration of the center of mass is $a_2 = \frac{P}{M}$.
• Rotational equation of motion about the center of mass: Since there is no friction, there is no torque about the center of mass. Therefore, the angular acceleration $\alpha_2 = 0$.
• This implies that the angular velocity remains constant throughout Phase 2, i.e., $\omega_2 = \omega_1$.
• The time taken for this phase, $t_2$, can be found from the angular displacement: $\Delta\theta_2 = \omega_1 t_2 \implies \pi = \omega_1 t_2$. Since $\omega_1 = v_1/R$, we have $t_2 = \frac{\pi}{\omega_1} = \frac{\pi R}{v_1}$.
• The distance moved by the center of mass in Phase 2, $x_2$, can be found using kinematics: $x_2 = v_1 t_2 + \frac{1}{2}a_2 t_2^2$. Substitute $t_2 = \frac{\pi R}{v_1}$ and $a_2 = \frac{P}{M}$: $x_2 = v_1 \left(\frac{\pi R}{v_1}\right) + \frac{1}{2}\left(\frac{P}{M}\right)\left(\frac{\pi R}{v_1}\right)^2$ $x_2 = \pi R + \frac{1}{2}\left(\frac{P}{M}\right)\frac{\pi^2 R^2}{v_1^2}$.
• From Phase 1, we found $v_1^2 = \frac{32P}{3M}$. From this, we can express $\frac{P}{M}$: $\frac{P}{M} = \frac{3v_1^2}{32}$.
• Substitute this expression for $\frac{P}{M}$ into the equation for $x_2$: $x_2 = \pi R + \frac{1}{2}\left(\frac{3v_1^2}{32}\right)\frac{\pi^2 R^2}{v_1^2}$ $x_2 = \pi R + \frac{3\pi^2 R^2}{64}$.
• Now substitute the value of $R = \frac{8}{\pi}$ cm: $x_2 = \pi \left(\frac{8}{\pi}\right) + \frac{3\pi^2 \left(\frac{8}{\pi}\right)^2}{64}$ $x_2 = 8 + \frac{3\pi^2 \left(\frac{64}{\pi^2}\right)}{64}$ $x_2 = 8 + \frac{3 \times 64}{64}$ $x_2 = 8 + 3 = 11$ cm.

Total distance moved by the center of the disc: The total distance moved by the center of the disc when it completes one rotation is the sum of distances from Phase 1 and Phase 2. $X_{total} = x_1 + x_2 = 8 \text{ cm} + 11 \text{ cm} = 19 \text{ cm}$.

Explanation of the solution:

1. Calculate Radius: From the given circumference, find the radius $R = C/(2\pi) = 16/(2\pi) = 8/\pi$ cm.
2. Phase 1 (Rough, No Slip): The first half rotation (from C to A) occurs on the rough surface where there's no slipping.
   • Distance moved by center of mass $x_1 = R \times (\text{angle}) = R\pi = (8/\pi)\pi = 8$ cm.
   • Using Newton's second law for translation ($P-f=Ma_1$) and rotation ($fR = I\alpha_1$), along with the no-slip condition ($a_1=R\alpha_1$ and $I=MR^2/2$), find the acceleration $a_1 = 2P/(3M)$.
   • Calculate the square of the velocity at the end of Phase 1: $v_1^2 = 2a_1 x_1 = 2(2P/(3M))(8) = 32P/(3M)$.
3. Phase 2 (Smooth, Slipping): The second half rotation (from A to C) occurs on the smooth surface, so friction is zero.
   • Linear acceleration $a_2 = P/M$ (since $f=0$).
   • Angular acceleration $\alpha_2 = 0$ (since no torque). This means angular velocity $\omega_2 = \omega_1$ (constant).
   • Time taken for this half rotation $t_2 = \text{angle}/\omega_1 = \pi/\omega_1 = \pi R/v_1$.
   • Distance moved by center of mass $x_2 = v_1 t_2 + (1/2)a_2 t_2^2$.
   • Substitute $t_2$ and $a_2$: $x_2 = \pi R + (1/2)(P/M)(\pi R/v_1)^2$.
   • Substitute $P/M = 3v_1^2/32$ (derived from $v_1^2$ expression in Phase 1): $x_2 = \pi R + (1/2)(3v_1^2/32)(\pi^2 R^2/v_1^2) = \pi R + 3\pi^2 R^2/64$.
   • Substitute $R=8/\pi$: $x_2 = \pi(8/\pi) + 3\pi^2(8/\pi)^2/64 = 8 + 3\pi^2(64/\pi^2)/64 = 8 + 3 = 11$ cm.
4. Total Distance: Sum the distances from both phases: $X_{total} = x_1 + x_2 = 8 + 11 = 19$ cm.

Subject: Physics Chapter: Rotational Motion (or System of Particles and Rotational Motion) Topic: Rolling Motion (or Combined Translation and Rotation) Difficulty Level: Medium Question Type: Integer