alleryStart learning →Torque on a body produces angular acceleration. For tau = 20 N m and I = 5 kg m², alpha equals:
- 4 rad/s²
- 100 rad/s²
- 20 rad/s²
- 5 rad/s²
Show answer and solution
Correct answer: A. 4 rad/s²
tau = I × alpha → alpha = tau/I = 20/5 = 4 rad/s².
CG PET Systems of Particles and Rotational Motion — practice questions — page 2
Explore 64 free Systems of Particles and Rotational Motion questions in CG PET Physics. Open each question or exercise to read its content, reference answers and available explanation.
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Showing questions 31–60 of 64.
Torque on a body produces angular acceleration. For tau = 20 N m and I = 5 kg m², alpha equals:Read the answer and worked solution →A 50 kg flywheel of radius 0.5 m rotates at 60 rev/min. Angular momentum (I = MR²/2):Read the answer and worked solution →A solid sphere rolls down without slipping. Fraction of KE that is rotational:Read the answer and worked solution →A wheel rotating at 50 rad/s decelerates uniformly to rest in 10 s. Total angular displacement:Read the answer and worked solution →Angular acceleration alpha = 4 rad/s² acts for 5 s on a wheel starting from rest. Number of revolutions:Read the answer and worked solution →A solid cylinder (I = MR²/2) of mass 2 kg, radius 0.1 m, rolls without slipping at 6 m/s. Total kinetic energy:Read the answer and worked solution →For a solid cylinder rolling without slipping, the friction force at the contact point:Read the answer and worked solution →Moment of inertia of an annular disc (inner radius r1, outer r2) about perpendicular axis through center:Read the answer and worked solution →Race four rolling bodies downhillRead the answer and worked solution →Follow a spinning skater's energyRead the answer and worked solution →What makes a body rigidRead the answer and worked solution →Rigid bodies and their motionRead the answer and worked solution →Rotation about a fixed axisRead the answer and worked solution →Odd one out: rotationRead the answer and worked solution →Angular velocity is a vectorRead the answer and worked solution →Linear velocity in rotationRead the answer and worked solution →The cross productRead the answer and worked solution →Vector operation to resultRead the answer and worked solution →The centre of massRead the answer and worked solution →Momentum of a systemRead the answer and worked solution →Defining the centre of massRead the answer and worked solution →Quantity to definitionRead the answer and worked solution →Torque and its other nameRead the answer and worked solution →Angular momentumRead the answer and worked solution →Odd one out: system quantitiesRead the answer and worked solution →Torque, in wordsRead the answer and worked solution →Which is largerRead the answer and worked solution →Same axis, same omegaRead the answer and worked solution →Translation versus rotationRead the answer and worked solution →Motion to descriptionRead the answer and worked solution →
Worked practice questions
A 50 kg flywheel of radius 0.5 m rotates at 60 rev/min. Angular momentum (I = MR²/2):
- 20 kg m²/s
- 314 kg m²/s
- 39.3 kg m²/s
- 3.14 kg m²/s
Show answer and solution
Correct answer: C. 39.3 kg m²/s
We have I = (1/2) × 50 × 0.5².
So I = 6.25 kg m².
omega = 60 rev/min = 60 × 2π/60 = 2π rad/s ≈ 6.283. L = I × omega = 6.25 × 6.283 ≈ 39.3 kg m²/s.
A solid sphere rolls down without slipping. Fraction of KE that is rotational:
- 5/7
- 2/5
- 1/2
- 2/7
Show answer and solution
Correct answer: D. 2/7
KE_rot/KE_total = (1/2) I omega² / [(1/2)Mv² + (1/2)I omega²] = (I/MR²) / (1 + I/MR²) (using v = omega R).
For sphere I/MR² = 2/5: ratio = (2/5)/(1 + 2/5) = (2/5)/(7/5) = 2/7. So 2/7 of KE is rotational.
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