Group1_RotationalMotion_Reviewer

Rotational Kinematics

Terminologies

• Angular Displacement (θ):

  • Angle an object rotates around a fixed axis, measured in radians (rad).

• Angular Velocity (ω):

  • Rate of change of angular displacement concerning time, measured in radians per second (rad/s).

• Angular Acceleration (α):

  • Rate of change of angular velocity concerning time, measured in radians per second squared (rad/s²).

Formulas

• Angular Velocity (ω) with Time:

  • ( \omega = \omega_0 + \alpha t )

• Angular Displacement (θ) with Time:

  • ( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 )

• Final Angular Velocity (ω) without Time:

  • ( \omega^2 = \omega_0^2 + 2 \alpha \theta )

• Average Angular Velocity (ω_avg):

  • ( \omega_{avg} = \frac{\omega_0 + \omega}{2} )

• Linear Velocity in relation to Angular Velocity:

  • ( v = r \cdot \omega )

• Object Revolution:

  • Total angular displacement regarding revolution: (\text{Rotations} = \frac{\theta}{2\pi} )

  • ( \theta = 2\pi N )

Sample Problems

Example 1:

• A bicycle slows down uniformly from an initial velocity of 8.40 m/s to rest over 115m with a wheel diameter of 68.0 cm. Determine:

  • (a) Initial angular velocity:

    • ( , v = r \cdot \omega )

    • ( \omega = \frac{v}{r} = \frac{8.40 m/s}{0.340 m} = 24.7 rad/s )

  • (b) Total revolutions before coming to rest:

    • Circumference ( C = 2\pi r = 2\pi(0.340 m) = 2.137 m )

    • ( N = \frac{d}{C} = \frac{115 m}{2.137 m} = 53.8 revolutions )

  • (c) Angular acceleration:

    • ( \omega^2 = \omega_0^2 + 2\alpha\theta )

    • ( \alpha = \frac{0 - (24.7)^2}{2(338)} = -0.902 rad/s² )

  • (d) Time taken to stop:

    • ( t = \frac{\omega_0}{\alpha} = \frac{24.7}{0.902} = 27.4 s )

Example 2:

• Yods riding a motorcycle with a rear wheel radius of 0.4m accelerates from rest to 20 m/s in 8 seconds. Determine:

  • (a) Angular acceleration:

    • Linear acceleration ( a = \frac{v - v_0}{t} = \frac{20 m/s - 0}{8 s} = 2.5 m/s² )

    • ( \alpha = \frac{a}{r} = \frac{2.5 m/s²}{0.4 m} = 6.25 rad/s² )

  • (b) Number of rotations ( \theta ):

    • ( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 = 0 + \frac{1}{2}(6.25)(8^2) = 200 radians )

    • ( N = \frac{200}{2\pi} = 31.85 )

  • (c) Final angular velocity:

    • ( \omega_f = \frac{v_f}{r} = \frac{20 m/s}{0.4 m} = 50 rad/s )

Rotational Dynamics

Definition of Terms

• Torque (τ):

  • Causes angular acceleration, the rotational equivalent of force in linear motion.

  • Depends on:

    • Magnitude of the applied force.

    • Perpendicular distance from the axis of rotation (lever arm).

  • Formula of Torque:

    • ( \tau = r \cdot F \sin \theta )

• Conditions for Zero Torque:

  • If the force passes through the axis of rotation (lever arm is zero).

  • If the force acts along the line of the lever arm (no perpendicular component).

• Units of Torque:

  • SI: Newton-meters (N·m)

  • English: Foot-pounds (ft·lb)

Net Torque and Angular Acceleration

• Angular acceleration is proportional to the net torque applied ( (\alpha = \frac{\tau}{I})), connecting Newton's second law for linear and rotational motion.

Example of Torque Application: Biceps Torque

• Case (a): Vertically applied force (90° to lever arm) yields maximum torque.

• Case (b): Force applied at an angle produces less torque:

  • ( \tau = r F \sin(θ) )

• Understanding torque is vital in biomechanics/tool design.

Angular Momentum

Linear Momentum

• ( p = mv ), quantity of motion of an object in straight motion.

  • In isolated systems, linear momentum is conserved.

Angular Momentum

• ( L = Iω ), describes motion around an axis.

  • A product of mass moment of inertia and angular velocity.

Conservation of Angular Momentum

• In absence of external torque, total angular momentum remains constant:

  • ( \Sigma τ = 0 )

  • ( L_0 = L_f )

Sample Problems

1. Two cylindrical plates MA (6.0 kg, 0.60 m) and MB (9.0 kg) in a clutch system rotating together. What is the final angular velocity?

   • ( I_0 \omega_0 = I_f \omega_f ) leads to ( \omega_f = ( \frac{m_A}{m_A + m_B} )\omega_0 = ( \frac{6.0}{15.0} )(7.2 rad/s) = 2.88 rad/s )

2. Merry-Go-Round Work Problem:

   • For a merry-go-round to accelerate from rest to 1 revolution in 7 s, net work required is calculated using work-energy relationship and moment of inertia descriptions.