Velocity and Acceleration Analysis Questions and Answers

Answer: C

The Coriolis acceleration is 1.00 m s⁻².

A point sliding along a rotating link has a Coriolis component of acceleration of magnitude \(2\omega v\), perpendicular to the link:

\[ \begin{aligned} a_{c} &= 2\omega v \\ &= 2 \times 1.0 \times 0.5 \\ &= 1.0\ \text{m s}^{-2} \end{aligned} \]

0.50 is what you get by leaving out the factor 2, which is the usual slip. 0.25 halves the product again, and 4.00 has no basis in the formula.

Answer: B

The follower turns at 0.1 rad/s.

The common normal at the point where the link touches the disc is perpendicular to the link and passes through the disc's geometric centre. In the position shown, that centre lies on the line joining the two fixed pivots, so it is the instantaneous centre shared by the disc and the follower. Measured from the pivots, it is 5 mm from the disc pivot and 45 + 5 = 50 mm from the follower pivot. By the angular velocity ratio theorem:

\[ \begin{aligned} \frac{\omega_{f}}{\omega_{d}} &= \frac{5}{50} \\ \omega_{f} &= 0.1 \times 1 = 0.1\ \text{rad/s} \end{aligned} \]

10.0 is the same ratio inverted (50/5), which would make the follower turn faster than the disc driving it. 0.05 and 5.0 do not correspond to the distances in the figure.

Answer: D

The angular velocity of the bar is \(\omega_{2} = 0.25\omega_{1}\).

Rolling without slipping makes the contact point R the instantaneous centre of the wheel. Take R as the origin: O is at (0, 3) and P, 2 m from O along the horizontal through O, is at (2, 3). So

\[ \begin{aligned} v_{P} &= \omega_{1}\,RP = \omega_{1}\sqrt{2^{2} + 3^{2}} \\ &= \sqrt{13}\,\omega_{1} \end{aligned} \]

and \(v_{P}\) is perpendicular to RP. Q, 8 m to the right of P, is at (10, 0) and slides along the floor, so its velocity is horizontal. The instantaneous centre I of the bar lies on the normal to each velocity: on the line RP extended, and on the vertical through Q. The line RP is \(y = 1.5x\), so at \(x = 10\) it gives I = (10, 15).

\[ \begin{aligned} IP &= \sqrt{8^{2} + 12^{2}} = 4\sqrt{13} \\ \omega_{2} &= \frac{v_{P}}{IP} \\ &= \frac{\sqrt{13}\,\omega_{1}}{4\sqrt{13}} \\ &= 0.25\,\omega_{1} \end{aligned} \]

The ratio is fixed by the geometry alone: the bar's centre I is four times as far from P as the wheel's centre R is, so the bar turns at a quarter of the wheel's rate. \(\omega_{2} = \omega_{1}\) would need the two centres equally far from P, and \(2\omega_{1}\) and \(0.5\omega_{1}\) do not match the distances in the figure.

Answer: A

The angular acceleration is the first expression, with \(n^{2} - 1\) over \((n^{2} - \sin^{2}\theta)^{3/2}\).

With crank radius r, connecting rod length l and \(n = l/r\), the connecting rod's angle \(\beta\) satisfies \(\sin\beta = \sin\theta / n\). Differentiating twice with respect to time at constant crank speed \(\omega\) gives

\[ \begin{aligned} \alpha_{c} &= -\omega^{2} \sin\theta \\ &\quad \times \frac{n^{2} - 1}{(n^{2} - \sin^{2}\theta)^{3/2}} \end{aligned} \]

The options differ only in the numerator sign or the power in the denominator. The \(3/2\) power comes from differentiating \((n^{2} - \sin^{2}\theta)^{-1/2}\), so powers of \(1/2\) or \(5/2\), or a numerator \(n^{2} + 1\), are wrong.

Answer: A

The acceleration at inner dead centre is \(rw^{2}\left(1 + \frac{1}{n}\right)\).

The piston acceleration in a slider crank mechanism, with \(n = l/r\), is approximately

\[ \begin{aligned} a &= r\omega^{2}\cos\theta \\ &\quad + \frac{r\omega^{2}\cos 2\theta}{n} \end{aligned} \]

At inner dead centre \(\theta = 0\), so both cosines equal 1 and \(a = r\omega^{2}(1 + 1/n)\). At outer dead centre, \(\theta = 180^\circ\), it becomes \(r\omega^{2}(1 - 1/n)\) in magnitude, which is the option with the minus sign. The other two do not follow from the formula.

Answer: C

The piston acceleration is \(\omega^{2}r\left(\cos\theta + \frac{\cos 2\theta}{n}\right)\), where \(n = l/r\).

For a crank of radius r turning at \(\omega\) with connecting rod length l, the piston displacement is approximately \(x = r\left[(1 - \cos\theta) + \frac{\sin^2\theta}{2n}\right]\). Differentiating once gives the velocity and twice gives the acceleration:

\[ \begin{aligned} v &= \omega r\left(\sin\theta + \frac{\sin 2\theta}{2n}\right) \\ a &= \omega^2 r\left(\cos\theta + \frac{\cos 2\theta}{n}\right) \end{aligned} \]

The sine forms belong to the velocity expression, not the acceleration, and the minus signs are wrong: at \(\theta = 0\) the acceleration is greatest, \(\omega^2 r(1 + 1/n)\), which needs a plus.

Answer: C

The velocity is at right angle to the link.

A rigid link cannot change length, so neither end can have any velocity relative to the other along the line joining them. The only relative motion possible is rotation about the other end, and the velocity of a point rotating about another point is perpendicular to the line joining them.

This is the basis of every velocity diagram: the relative velocity of B with respect to A is drawn perpendicular to AB. A component parallel to the link, or at 45° to it, would stretch or shorten the link, and −180° is simply a direction along the link.

Answer: C

The three instantaneous centres lie on a straight line.

Kennedy's theorem, the theorem of three centres, states that if three bodies have plane motion relative to one another, their three instantaneous centres lie on one straight line. It is used to locate instantaneous centres that cannot be found by inspection.

The centres are three distinct points in general, so they cannot coincide at a single point, and the theorem says nothing about a curved path or a centre of curvature.

Answer: A

The total number of instantaneous centres is \(\frac{n(n-1)}{2}\).

Every pair of links in a mechanism has one instantaneous centre, so the number of centres is the number of ways of choosing two links from n:

\[ N = \binom{n}{2} = \frac{n(n-1)}{2} \]

A four-bar chain, for example, has \(\frac{4 \times 3}{2} = 6\) instantaneous centres. The options n, \(\frac{n}{2}\) and \(\frac{n-1}{2}\) do not count the pairs; for four links they give 4, 2 and 1.5, and a count of centres cannot be fractional.

Answer: A

The number of instantaneous centres is \(\frac{n(n-1)}{2}\).

Every pair of links in a mechanism has one instantaneous centre. The number of pairs that can be formed from n links is the number of combinations of n taken two at a time.

\[ N = \frac{n(n-1)}{2} \]

A four-bar chain therefore has 6 instantaneous centres. \(n(n-1)\) counts each pair twice, and \((n-1)\) and \((n-1)/2\) undercount.