Rotational mechanics for AP Physics C: Mechanics

Reading cue. The dashed line exists to fix that both start at the same height h; the labels under the two ramps — rolls without slipping, frictionless — are the whole difference between the cases.

1. aIndicate Indicate whether the gravitational potential energy converted by the sphere on its way to the floor is greater than, less than, or equal to that converted by the block. No justification is required.

   recall1 mark

   Check answer · 1 mark

   1. equal: both have mass M and both fall through the same height h from the dashed line, so each converts Mgh

2. bDetermine Determine what fraction of the sphere's total kinetic energy at the foot of the ramp is rotational.

   routine3 marks

   Check answer · 3 marks

   1. rolling without slipping gives ω = v/R, so the rotational term is ½(2/5)MR²(v/R)² = (1/5)Mv²
   2. total kinetic energy = ½Mv² + (1/5)Mv² = (7/10)Mv²
   3. fraction rotational = (1/5)/(7/10) = 2/7, about 0.29

3. cDetermine Both objects cover the same distance along the face of their identical ramps, starting from rest with uniform acceleration. Determine the ratio of the time the sphere takes to reach the floor to the time the block takes.

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   Check answer · 4 marks

   1. for uniform acceleration from rest the average speed is half the final speed, so the face length L gives t = 2L/v for each object
   2. L is the same for both, so the ratio of times is the inverse ratio of the final speeds: t_sphere/t_block = v_block/v_sphere
   3. v_block = √(2gh) and v_sphere = √(10gh/7), so the ratio is √(2 ÷ 10/7) = √(7/5)
   4. t_sphere/t_block = √1.4 = 1.18, so the sphere takes about 18% longer

4. dDerive The sphere on the left ramp is replaced by a hollow spherical shell of the same mass and radius, for which I = (2/3)MR², released from rest on the same dashed line. Derive its speed at the floor and explain where it ranks against the two speeds the figure compares.

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   Check answer · 4 marks

   1. energy conservation with ω = v/R: Mgh = ½Mv² + ½(2/3)MR²(v/R)² = (5/6)Mv²
   2. v_shell = √(6gh/5) = 1.10√(gh), against 1.20√(gh) for the solid sphere and 1.41√(gh) for the block
   3. the shell is slowest: v_shell/v_sphere = √(0.84) = 0.92, so it is about 8% slower than the sphere
   4. all of the shell's mass sits at the rim, giving the largest moment of inertia for the same M and R, so the largest share of the Mgh goes into rotation and the least into translation

Reading cue. The dimension line runs from the axis symbol to the far end, so L is the whole rod and the axis is at the end, not the centre; the weight acts a distance L/2 from that axis.

1. aIndicate Indicate the point on the rod through which its weight acts, and state its distance from the axis marked on the figure.

   recall2 marks

   Check answer · 2 marks

   1. the midpoint of the rod, since the rod is uniform
   2. a distance L/2 from the marked axis

2. bDerive Taking the moment of inertia about the axis drawn to be ML²/3, derive the moment of inertia about a parallel axis through the centre of the rod.

   routine3 marks

   Check answer · 3 marks

   1. parallel-axis theorem used in the form I_axis = I_cm + Md²
   2. d = L/2, so Md² = ML²/4
   3. I_cm = ML²/3 − ML²/4 = ML²/12

3. cDetermine Determine every position along the rod at which a parallel axis would give a moment of inertia exactly twice that about the centre, giving each as a distance from the axis drawn.

   demanding4 marks

   Show a hint

   There is more than one such axis.

   Check answer · 4 marks

   1. condition I_cm + Md² = 2I_cm, so Md² = I_cm
   2. d² = L²/12, giving d = L/(2√3) = 0.289L from the centre of mass
   3. two axes, at 0.5L ± 0.289L from the drawn axis
   4. 0.21L and 0.79L from the drawn axis, both lying on the rod

4. dDerive The rod is replaced by one of the same mass M and the same length L, but with a linear density that increases with distance x from the drawn axis as λ = λ₀x/L. Derive its moment of inertia about that axis and compare it with ML²/3.

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   Check answer · 4 marks

   1. M = ∫₀^L (λ₀x/L) dx = λ₀L/2, so λ₀ = 2M/L
   2. dI = x² dm with dm = λ dx = (2M/L²)x dx
   3. I = ∫₀^L (2M/L²)x³ dx = ML²/2
   4. 1.5 times the uniform value, because mass has been shifted towards the far end where the lever arm x is largest

Reading cue. The dashed outline is where the rod ends up, not a second rod: what falls is the centre of mass at L/2 from the pivot, and it falls L/2, not L.

1. aIndicate Indicate the point whose fall supplies the energy released as the rod swings to the dashed position, and state how far that point drops.

   recall2 marks

   Check answer · 2 marks

   1. the centre of mass, at the midpoint of the rod
   2. it drops a vertical distance L/2, not L

2. bDetermine Determine the angular acceleration of the rod at the instant it reaches the dashed vertical position.

   routine3 marks

   Check answer · 3 marks

   1. at the vertical the line of action of the weight passes through the pivot
   2. the moment arm, and hence the torque about the pivot, is zero
   3. α = 0 at that instant, even though the rod is turning fastest there

3. cDetermine Determine the horizontal and vertical components of the force the pivot exerts on the rod at the instant of release, in terms of M and g.

   demanding4 marks

   Check answer · 4 marks

   1. α = (MgL/2)/(ML²/3) = 3g/2L at release
   2. centre of mass acceleration a = αL/2 = 3g/4 downwards
   3. Newton's second law vertically: Mg − N = M(3g/4), so N = Mg/4 upwards
   4. ω = 0 at release, so no centripetal force is needed and the horizontal component is zero

4. dDerive A small coin rests on the rod a distance x from the pivot and is released with it. Derive the least value of x for which the coin loses contact with the rod immediately, and describe what is seen for a coin placed closer to the pivot.

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   Check answer · 4 marks

   1. the point of the rod under the coin has downward acceleration a = αx = 3gx/2L at release
   2. a coin acted on by gravity alone can accelerate downwards only at g, since the rod cannot pull it down
   3. contact is lost when 3gx/2L > g, that is x > 2L/3
   4. for x < 2L/3 the rod pushes up on the coin, the two accelerate together and the coin stays on the rod

Reading cue. The string leaves the small hub, not the wheel rim, so r is the moment arm and the block's acceleration is αr; the wheel's own radius enters nowhere in the mechanics.

1. aIndicate Indicate whether the moment arm of the string tension about the axle is the hub radius r or the radius of the large wheel, and state the relation between the block's linear acceleration and the wheel's angular acceleration.

   recall3 marks

   Check answer · 3 marks

   1. the hub radius r, because the string leaves the hub tangentially
   2. a = αr
   3. the outer radius of the wheel does not appear in the relation, because no force is applied at the rim

2. bDetermine The block accelerates downwards with acceleration a. Determine the tension in the string in terms of m, g and a, and explain why the tension must be smaller than the weight of the block.

   routine3 marks

   Check answer · 3 marks

   1. Newton's second law on the block: mg − T = ma
   2. T = m(g − a)
   3. for a > 0 this requires T < mg; if T equalled mg the block could not accelerate downwards at all

3. cCalculate In one trial the hub radius is r = 4.0 cm and the block has mass 0.50 kg. From rest, the block falls 1.20 m in 3.0 s. Ignoring axle friction, calculate the moment of inertia of the wheel. Take g = 9.8 m s⁻².

   demanding4 marks

   Check answer · 4 marks

   1. a = 2h/t² = 2(1.20)/3.0² = 0.267 m s⁻²
   2. T = m(g − a) = 0.50(9.8 − 0.267) = 4.77 N
   3. α = a/r = 0.267/0.040 = 6.67 rad s⁻²
   4. I = Tr/α = (4.77 × 0.040)/6.67 = 2.9 × 10⁻² kg m²

4. dDetermine After the block lands, the string leaves the hub and the wheel is left to spin freely; it comes to rest 12 s later. Determine the frictional torque in the axle, determine a corrected moment of inertia, and determine the percentage error in the value found in the previous part.

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   Check answer · 5 marks

   1. speed at landing v = at = 0.80 m s⁻¹, so ω₀ = v/r = 20 rad s⁻¹
   2. while coasting, friction alone acts: α_f = ω₀/t = 20/12 = 1.67 rad s⁻² and τ_f = Iα_f
   3. during the fall Tr − τ_f = Iα, giving 0.191 = I(6.67 + 1.67)
   4. I = 2.3 × 10⁻² kg m² and τ_f = 3.8 × 10⁻² N m
   5. the frictionless value is 25% too high, because torque actually spent against friction was credited to the wheel's inertia