Plane Motion of Rigid Bodies Questions and Answers

Answer: A

The flywheel is a uniform solid disc, so its moment of inertia about its axis is \(I = \tfrac{1}{2}mr^2 = \tfrac{1}{2}(10)(0.1)^2 = 0.05\ \text{kg m}^2\). The speed is 300 rpm, so \(\omega = 2\pi \times 300/60 = 31.42\) rad/s. The rotational kinetic energy is

The nearest option is 24.65 J. The thickness of 15 mm is not needed since the mass is given. The larger options would come from using the wrong moment of inertia, such as that of a ring (\(mr^2\)), which would double the answer.

Answer: A

The ratio is \(\sqrt{3/4}\).

Rolling without slipping, energy conservation gives \(mgh = \tfrac{1}{2}mv^{2}(1 + I/mR^{2})\).

\[ \begin{aligned} \text{A (ring, } I = mR^{2}) &: v_{A}^{2} = gh \\ \text{B (disk, } I = \tfrac{1}{2}mR^{2}) &: v_{B}^{2} = \tfrac{4}{3}gh \\ \frac{v_{A}}{v_{B}} &= \sqrt{\tfrac{3}{4}} \end{aligned} \]

Disk A, with its mass at the rim, stores more of its energy as rotation and so arrives slower. \(\sqrt{3/2}\) and \(\sqrt{2}\) put A ahead, and 1 would hold only if both had the same moment of inertia.

Answer: D

The acceleration of R is 8∠217° m/s².

In plane rigid-body motion, the acceleration of one point relative to another is \(\vec{a}_{B/A} = \vec{\alpha}\times\vec{r}_{AB} - \omega^{2}\vec{r}_{AB}\). Its magnitude is \(|r_{AB}|\sqrt{\alpha^{2}+\omega^{4}}\), and it makes the same fixed angle with \(\vec{r}_{AB}\) for every pair of points.

From P to R the distance is 20 and the direction is that of (16, -12), at -36.87° (323.13°). The given relative acceleration is 10 at 180°, so

\[ \begin{aligned} \sqrt{\alpha^{2}+\omega^{4}} &= 10/20 = 0.5 \\ \text{angle offset} &= 180^\circ - 323.13^\circ = -143.13^\circ \end{aligned} \]

Since Q has zero acceleration, \(\vec{a}_{R} = \vec{a}_{R/Q}\). From Q to R the distance is 16 at 0°, so \(|a_{R}| = 16 \times 0.5 = 8\) at 0° - 143.13° = 216.87°, about 217°.

8∠233° has the right size but the wrong direction, and the options with magnitude 10 simply copy the given relative acceleration.

Answer: A

The speed of Q is 5 m/s.

Put the corner at the origin, with P at (x, 0) on the floor and Q at (0, y) on the wall. The rod's length fixes \(x^{2} + y^{2} = 1\). Differentiating with respect to time:

\[ \begin{aligned} x\dot{x} + y\dot{y} &= 0 \\ |\dot{y}| &= \frac{x}{y}\,\dot{x} = \cot\theta \times U \end{aligned} \]

At θ = 45°, x = y, so cot θ = 1 and V = U = 5 m/s, directed down the wall.

The same result follows from the instantaneous centre, which lies at the intersection of the normals to the two velocities, equidistant from P and Q when θ = 45°. The rod's length does not affect the speed ratio, and 6, 8 and 10 have no basis in this geometry.

Answer: D

The acceleration of the contact point is \(r\omega^{2}\), directed towards the centre of the disc.

The centre moves horizontally with acceleration \(r\alpha\). Relative to the centre, the contact point has a tangential acceleration \(r\alpha\) and a centripetal acceleration \(r\omega^{2}\). At the bottom of the disc the tangential part points opposite to the centre's acceleration, so the two \(r\alpha\) terms cancel, leaving only the centripetal term, which points straight up.

'Zero' is the velocity of the contact point in rolling without slipping, not its acceleration. \(r\alpha\) is the centre's acceleration, and the square-root expression adds the components that in fact cancel.

Answer: A

The velocities are \(r\omega\,\hat{i}\) at O and \(0\,\hat{i}\) at A.

In rolling without slipping, the contact point A has zero velocity, because it does not slide on the ground. The disc turns clockwise, as the arrow shows, so it rolls to the right, in the \(+\hat{i}\) direction.

The centre is a distance \(r\) above the instantaneous centre A, so its speed is \(r\omega\), directed along \(+\hat{i}\). \(-r\omega\,\hat{i}\) would be anticlockwise rolling. The options giving A a velocity of \(\pm r\omega\) contradict the no-slip condition.

Answer: A

The velocity of A is \((1 - 2\sqrt{3})\)i + 2j m/s.

The rod makes 60° with the x-axis and is 2 m long, so the position of A relative to B is

\[ \begin{aligned} \mathbf{r}_{A/B} &= 2\cos 60^{\circ}\,\mathbf{i} + 2\sin 60^{\circ}\,\mathbf{j} \\ &= \mathbf{i} + \sqrt{3}\mathbf{j} \end{aligned} \]

It turns anticlockwise, \(\boldsymbol{\omega} = 2\mathbf{k}\):

\[ \begin{aligned} \mathbf{v}_{A} &= \mathbf{v}_{B} + \boldsymbol{\omega} \times \mathbf{r}_{A/B} \\ &= \mathbf{i} + 2\mathbf{k} \times (\mathbf{i} + \sqrt{3}\mathbf{j}) \\ &= \mathbf{i} + 2\mathbf{j} - 2\sqrt{3}\mathbf{i} \end{aligned} \]

\(-2\sqrt{3}\)i + 2j leaves out the velocity of B, and the options with \(-2\)j take the rotation as clockwise.

Answer: B

The angular acceleration is \(\frac{F}{2mc}\).

On a smooth floor the assembly is free. Its centre of mass is at O, and the moment of inertia about O is that of four masses at distance c:

\[ \begin{aligned} I_{O} &= 4mc^{2} \\ \tau_{O} &= 2F \cdot c \\ \alpha &= \frac{2Fc}{4mc^{2}} = \frac{F}{2mc} \end{aligned} \]

The force also accelerates the centre of mass, at \(2F/4m\), but that does not change the rotation about the centre of mass. \(F/(mc)\) uses only two of the masses in the inertia, and \(2F/(3mc)\) and \(F/(3mc)\) do not match the inertia \(4mc^{2}\) of four equal masses.

Answer: D

The speed of the centre after impact is \(0.9v\).

The impulse at S passes through S, and there is no impulse from PQ, so angular momentum about S is conserved during the impact. Before impact the centre moves horizontally at \(v\), at height \(R - \frac{3R}{20} = \frac{17R}{20}\) above S, and the disk spins at \(\omega = v/R\):

\[ \begin{aligned} H_{S} &= mv\tfrac{17R}{20} + \tfrac{1}{2}mR^{2}\tfrac{v}{R} \\ &= \tfrac{27}{20}mvR \end{aligned} \]

After impact the disk pivots about S without slip or rebound, with moment of inertia about S of \(\frac{1}{2}mR^{2} + mR^{2} = \frac{3}{2}mR^{2}\):

\[ \begin{aligned} \tfrac{3}{2}mR^{2}\omega' &= \tfrac{27}{20}mvR \\ \omega' &= 0.9\tfrac{v}{R}, \quad v' = \omega' R = 0.9v \end{aligned} \]

Leaving out the spin term \(\frac{1}{2}mR^{2}\omega\) would give only about \(0.57v\); the disk's own rotation carries angular momentum into the impact, which is why the speed falls by only a tenth. 0.1v, 0.3v and 0.7v would all require angular momentum about S to be lost, which the stated conditions rule out.

Answer: B

The force must be \(\frac{mg}{2}\).

Both bars are horizontal with F acting downward at the free end, so gravity adds to F in the end-pinned case but has no moment about the centre pin in the second case.

\[ \begin{aligned} \alpha_1 &= \frac{2FL + mgL}{\tfrac{4}{3}mL^{2}} = \frac{3(2F + mg)}{4mL} \\ \alpha_2 &= \frac{FL}{\tfrac{1}{3}mL^{2}} = \frac{3F}{mL} \end{aligned} \]

using \(I = m(2L)^{2}/3\) about an end and \(I = m(2L)^{2}/12\) about the centre. Setting \(\alpha_1 = \alpha_2\) gives \(2F + mg = 4F\), so \(F = mg/2\).

Ignoring gravity would make the two accelerations unequal for any F, and mg, mg/4 and mg/6 do not satisfy the balance above.