NEB

Grade 11 Physics: Circular Motion

CDC Nepal Standard Textbook Module

National Examinations Board (NEB) - Grade 11 Physics

Unit 6: Circular Motion (Complete Study & Derivation Textbook)

Prepared in accordance with the official CDC Nepal Syllabus. Includes line diagrams, derivations, and board examination numerical solutions.

NEB Curriculum Guide

Unit 6: Circular Motion

A complete standard self-study resource covering angular kinematics, force dynamics, banked roads, conical pendulums, and vertical loops with complete NEB board exam proofs.

Section 6.1

Angular Displacement, Velocity, and Acceleration

When a particle moves along a circular path of radius $r$, its motion is called circular motion. To describe this motion, we use angular variables analogous to linear kinematics.

1. Angular Displacement ($\theta$)

The angle swept out by the radius vector at the center of the circular path in a given time interval.

$\theta = \frac{s}{r} \text{ radians}$

  • SI Unit: radian ($\text{rad}$)
  • Dimensions: $[M^0L^0T^0]$ (Dimensionless)

2. Angular Velocity ($\omega$)

The rate of change of angular displacement with respect to time.

$\omega = \frac{d\theta}{dt} = \frac{2\pi}{T} = 2\pi f$

  • SI Unit: $\text{rad/s}$
  • Dimensions: $[M^0L^0T^{-1}]$

3. Angular Acceleration ($\alpha$)

The rate of change of angular velocity with time.

$\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$

  • SI Unit: $\text{rad/s}^2$
  • Dimensions: $[M^0L^0T^{-2}]$
O Vector ω, α P(t) r v Counter-clockwise Motion

Direction of Axial Vectors (Right-Hand Thumb Rule)

Angular displacement ($\boldsymbol{\theta}$), velocity ($\boldsymbol{\omega}$), and acceleration ($\boldsymbol{\alpha}$) are axial vectors. Their direction is perpendicular to the plane of rotation.

"Curl the fingers of your right hand in the direction of circular rotation of the body; then the stretched thumb points along the axis of rotation in the direction of $\vec{\omega}$ and $\vec{\alpha}$."

Analogy Between Linear and Angular Kinematics

Linear Parameter Angular Parameter Connecting Relation
Linear Displacement ($s$) Angular Displacement ($\theta$) $s = r\theta$
Linear Velocity ($v$) Angular Velocity ($\omega$) $v = r\omega$
Tangential Acceleration ($a_t$) Angular Acceleration ($\alpha$) $a_t = r\alpha$
Mass ($m$) Moment of Inertia ($I$) $I = \sum m r^2$
Section 6.7

Conical Pendulum

A conical pendulum consists of a small heavy bob of mass $m$ tied to a light inextensible string of length $l$, suspended from a fixed point, such that the bob rotates in a horizontal circle at constant angular velocity.

S h l θ r m mg Ts Tscosθ Tssinθ

Force Resolution:

Let string length be $l$, inclination with vertical be $\theta$, and radius of path be $r = l \sin\theta$. Height $h = l \cos\theta$.

Tension $T_s$ in string resolves into two components:

  • $T_s \cos\theta$ (Vertical component): Balances downward weight $mg$.
  • $T_s \sin\theta$ (Horizontal component): Provides required centripetal force ($m\omega^2 r$).

Derivation of Time Period ($T$):

From force equilibrium equations:

Equation (1): $T_s \cos\theta = mg$
Equation (2): $T_s \sin\theta = m \omega^2 r = m \omega^2 (l \sin\theta)$

From Equation (2), dividing both sides by $\sin\theta$:

$$T_s = m \omega^2 l$$

Substitute $T_s$ in Equation (1):

$$(m \omega^2 l) \cos\theta = mg \implies \omega^2 = \frac{g}{l \cos\theta}$$

Since angular velocity $\omega = \frac{2\pi}{T}$:

$$\left(\frac{2\pi}{T}\right)^2 = \frac{g}{l \cos\theta} \implies \frac{2\pi}{T} = \sqrt{\frac{g}{l \cos\theta}}$$
$$T = 2\pi \sqrt{\frac{l \cos\theta}{g}} = 2\pi \sqrt{\frac{h}{g}}$$

Frequency ($f$): $f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g}{l \cos\theta}}$

String Tension ($T_s$): $T_s = \frac{mg}{\cos\theta}$

Limiting Case: If $\theta \to 0^\circ$, $\cos\theta \to 1$, reducing to simple pendulum formula $T = 2\pi \sqrt{\frac{l}{g}}$.

Section 6.8

Motion in a Vertical Circle

Consider a small body of mass $m$ tied to an inextensible string of length $r$ rotated in a vertical circle. Unlike horizontal circular motion, motion in a vertical circle involves variable speed and changing tension.

O L TL mg T TT mg M TM

Critical Conditions for Complete Looping:

  • At Highest Point (T):
    $T_T + mg = \frac{m v_T^2}{r}$
    For minimum critical velocity, $T_T \ge 0 \implies v_T \ge \sqrt{gr}$.
  • At Lowest Point (L):
    $T_L - mg = \frac{m v_L^2}{r}$
    Using conservation of energy, $v_L \ge \sqrt{5gr}$.
  • At Horizontal Point (M):
    $v_M \ge \sqrt{3gr}$

Complete Step-by-Step Derivation

1. Critical Velocity at Highest Point ($v_T$):

At top point $T$, both gravity $mg$ and string tension $T_T$ act downwards toward center $O$:

$$T_T + mg = \frac{m v_T^2}{r} \implies T_T = \frac{m v_T^2}{r} - mg$$

To complete the loop without string slackening, $T_T \ge 0$:

$$\frac{m v_T^2}{r} - mg \ge 0 \implies v_T^2 \ge gr \implies \mathbf{v_T(\text{min}) = \sqrt{gr}}$$

2. Critical Velocity at Lowest Point ($v_L$):

By Conservation of Mechanical Energy between lowest point $L$ ($h=0$) and top point $T$ ($h=2r$):

$$E_L = E_T \implies \frac{1}{2} m v_L^2 + 0 = \frac{1}{2} m v_T^2 + mg(2r)$$

Substitute minimum top velocity $v_T^2 = gr$:

$$\frac{1}{2} m v_L^2 = \frac{1}{2} m (gr) + 2 mgr = \frac{5}{2} mgr$$
$$v_L(\text{min}) = \sqrt{5gr}$$

3. Difference in String Tension ($T_L - T_T$):

At bottom: $T_L = \frac{m v_L^2}{r} + mg$. At top: $T_T = \frac{m v_T^2}{r} - mg$.

$$T_L - T_T = \left(\frac{m v_L^2}{r} + mg\right) - \left(\frac{m v_T^2}{r} - mg\right) = \frac{m}{r}(v_L^2 - v_T^2) + 2mg$$

Since $v_L^2 - v_T^2 = 4gr$ from energy conservation:

$$T_L - T_T = \frac{m}{r}(4gr) + 2mg = 6mg$$

The difference in tension between the lowest and highest points is always $6mg$, independent of the velocity.

Complete study module for NEB Grade 11 Physics - Unit 6: Circular Motion