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  1. Problems
  2. /A uniform disc of mass 16 kg and radius 0.40 m turns on a fixed axle through its centre…
MediumPhysicsHigh School (11–12)Mechanics › Rotational MotionNumericalBoard ExamAPIB

A uniform disc of mass 16 kg and radius 0.40 m turns on a fixed axle through its centre…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A uniform disc of mass 16 kg16\ \text{kg}16 kg and radius 0.40 m0.40\ \text{m}0.40 m turns on a fixed axle through its centre. A steady torque of 4.0 N m4.0\ \text{N m}4.0 N m acts on it from rest. Find

  1. its angular acceleration;
  2. the time it takes to reach 50 rad/s50\ \text{rad/s}50 rad/s;
  3. its kinetic energy then.

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

rad/s²
s
J
Need a reminder? The key ideaShowHide

For rotation about a fixed axis: τ=Iα\tau = I\alphaτ=Iα, kinetic energy 12Iω2\tfrac12 I\omega^221​Iω2, work τθ\tau\thetaτθ, power τω\tau\omegaτω. A rope on a wheel links the two motions: a=αRa = \alpha Ra=αR.

Moment of inertia: I=∑miri2I = \sum m_ir_i^2I=∑mi​ri2​. Ring MR2MR^2MR2; disc 12MR2\tfrac12 MR^221​MR2; rod 112ML2\tfrac{1}{12}ML^2121​ML2 about its middle; solid sphere 25MR2\tfrac25 MR^252​MR2. Parallel axes: I=Icm+Md2I = I_{cm} + Md^2I=Icm​+Md2. Flat bodies: Iz=Ix+IyI_z = I_x + I_yIz​=Ix​+Iy​.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. I=12×16 kg×(0.40 m)2=1.28 kg m2I = \tfrac12 \times 16\ \text{kg} \times (0.40\ \text{m})^2 = 1.28\ \text{kg m}^2I=21​×16 kg×(0.40 m)2=1.28 kg m2, τ=4.0 N m\tau = 4.0\ \text{N m}τ=4.0 N m.

Step 1. (a) α=4.0 N m1.28 kg m2=3.125 rad/s2\alpha = \dfrac{4.0\ \text{N m}}{1.28\ \text{kg m}^2} = 3.125\ \text{rad/s}^2α=1.28 kg m24.0 N m​=3.125 rad/s2.

Step 2. (b) t=50 rad/s3.125 rad/s2=16 st = \dfrac{50\ \text{rad/s}}{3.125\ \text{rad/s}^2} = 16\ \text{s}t=3.125 rad/s250 rad/s​=16 s.

Step 3. (c) K=12×1.28 kg m2×(50 rad/s)2=1600 JK = \tfrac12 \times 1.28\ \text{kg m}^2 \times (50\ \text{rad/s})^2 = 1600\ \text{J}K=21​×1.28 kg m2×(50 rad/s)2=1600 J.

Answer: (a) 3.125 rad/s23.125\ \text{rad/s}^23.125 rad/s2. (b) 16 s16\ \text{s}16 s. (c) 1600 J1600\ \text{J}1600 J.

Check. The torque's work: θ=12×3.125 rad/s2×(16 s)2=400 rad\theta = \tfrac12 \times 3.125\ \text{rad/s}^2 \times (16\ \text{s})^2 = 400\ \text{rad}θ=21​×3.125 rad/s2×(16 s)2=400 rad, and τθ=1600 J\tau\theta = 1600\ \text{J}τθ=1600 J. Same.

Watch out. A disc's moment of inertia is 12MR2\tfrac12 MR^221​MR2, not MR2MR^2MR2.

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  1. Dynamics of rotation about a fixed axisTorque, Moment of Inertia and Rotation About a Fixed Axis

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