Study Guide 11 - Mechanics of Biological Systems: Rotaional Motion

The bird travels in a circle because there is a net force toward the centre of the circle called the centripetal force.

The centripetal force which is given by:
\(\mathrm {f_c = mv^2/r}\)
where r is the radius of the circle and m is the mass of the object and v is its tangential velocity.

\(\mathrm {f_c = mv^2/r = 0.5(10^2)/20 = 2.5\; N}\)

The net force on the bird is the same at all points of the circle if the velocity is constant so the answer is the same as for part (a)

Since the net force on the bird is the same at every part of the circle, at the bottom the air must exert an upward lift force \(\mathrm {(f_L)}\) to provide this centripetal force and support the weight of the bird.

\(\mathrm {f_c = f_L - mg}\)

\(\mathrm{f_L = f_c + mg = 2.5 + 0.5(9.8) = 7.4\; N}\)
 

(B) \(\mathrm {F = mv^2/r}\)

The centripetal force is provided by the tension in the string. Using:

\(\mathrm {F = mv^2/r = 0.5 \times 20^2/1.5 = 133\; N}\)
 

Moments can be taken about any point but if \(\mathrm{T_2}\) is wanted which is the best point to choose in this problem?

Taking moments about \(\mathrm{O_1}\) makes the unknown \(\mathrm {T_1}\) disappear from the problem.

Take moments about \(\mathrm {O_1}\) and set their sum equal to zero.

Counter-clockwise torque produced by \(\mathrm {T_2}\) is \(\mathrm{T_2 \times 4.00\; \sin\; 60  = 3.464 \;T_2}\)  

See diagram below:

Clockwise torque produced by the weight of the scaffold (acting at its centre) and the weight of the man i.e.,
\(\mathrm {(750)(2.00) + (100)(9.8)(1.00) = 2480}\)
(Be careful to note that the weight of the scaffold in N is given but the mass of the man in kg is given)

\(\mathrm {\sum \tau = 0 \\ 3.464\; T_2 - 2480 = 0 \\ T_2 = 2480/3.464 = 716\; N}\)

Using moments, the value of T2 was found in Part (a)
The value was \(\mathrm {716\; N}\)

Now \(\mathrm{\sum F_x = 0}\) and \(\mathrm {\sum F_y = 0}\) can be used to find the other two unknowns \(\mathrm{T_1}\) and \(\mathrm {\theta.}\)

\(\mathrm{T_2\; \cos \;60 - T_1\; \sin \theta = 0 \;or \\ T_1 \; \sin \theta = T_2\; \cos \;60 = 358 \qquad [1]}\)

\(\mathrm{T_2\; \sin \;60 +  T_1\cos \theta - 100(9.8) - 750 = 0 \;or \\ T_1\cos \theta = - T_2\; \sin \;60 + 980 + 750 = 1110  \qquad  [2]}\)

Divide [1] by [2]

\(\mathrm{\tan \theta = 358/1110 =0.322 \\ \theta = 17.9^\circ \; {\color {red} {This\; is \;the\; answer\; to\; part\; (c)}}}\)

Use either [1] or [2] to find \(\mathrm{T_1}\)
Using [1]
\(\mathrm{T_1 = 358/\sin 17.9 = 358/0.307 = 1170\; N}\)

Using moments, the value of T2 was found in Part (a)
The value was \(\mathrm {716\; N}\)

Now \(\mathrm{\sum F_x = 0}\) and \(\mathrm {\sum F_y = 0}\) can be used to find the other two unknowns \(\mathrm{T_1}\) and \(\mathrm {\theta.}\)

\(\mathrm{T_2\; \cos \;60 - T_1\; \sin \theta = 0 \;or \\ T_1 \; \sin \theta = T_2\; \cos \;60 = 358 \qquad [1]}\)

\(\mathrm{T_2\; \sin \;60 +  T_1\cos \theta - 100(9.8) - 750 = 0 \;or \\ T_1\cos \theta = - T_2\; \sin \;60 + 980 + 750 = 1110  \qquad  [2]}\)

Divide [1] by [2]

\(\mathrm{\tan \theta = 358/1110 =0.322 \\ \theta = 17.9^\circ \; {\color {red} {This\; is \;the\; answer\; to\; part\; (c)}}}\)

Use either [1] or [2] to find \(\mathrm{T_1}\)
Using [1]
\(\mathrm{T_1 = 358/\sin 17.9 = 358/0.307 = 1170\; N}\)

Taking moments about A

\(\mathrm{Td - 400(5 \; \cos \; 53.1) - 1000(10\; \cos \;53.1) = 0}\)

\(\mathrm {T(6\; \sin \;30) = 400(5 \times 0.6) + 1000(10 \times 0.6)}\)

\(\mathrm{T(6 \times 0.5) = 1200 + 6000 = 7200}\)

\(\mathrm {T = 7200/3 = 2400\; N}\)

The relationship between speed, angular velocity and radius is:  \(\mathrm {v = r\omega}\)

\(\mathrm {3000\; rpm\; must\; be\; expressed\; in\; rad/s \\ 3000\; rpm = 3000(2\pi )/60 = 314\; rad/s \\ v = r\omega = 0.25 \times 314 = 78.5\; m/s}\)

Which is the appropriate equation to use in this problem? Note that all of the following are correct equations.

(A)    \(\mathrm {\omega = \omega_0 + \alpha t}\)

(B)    \(\mathrm {\omega = \omega_0 + \omega_0t + ½\alpha t^2}\)

(C)    \(\mathrm {\omega^2 = \omega_0{^2} + 2\alpha \theta}\)

In part (a) we calculated that the final angular velocity was 314 rad/s

\(\mathrm {\omega = \omega_0 + \alpha t}\)

\(\mathrm {314 = 0 + \alpha (3 \times 60)}\)

\(\mathrm {\alpha = 1.75\; rad/s^2}\)

 Which is the appropriate equation to use in this problem? Note that all of the following are correct equations.

(A)    \(\mathrm {\omega = \omega_0 + \alpha t}\)

(B)    \(\mathrm {\omega = \omega_0 + \omega_0t + ½\alpha t^2}\)

(C)    \(\mathrm {\omega^2 = \omega_0{^2} + 2\alpha \theta}\)

\(\mathrm {\omega^2 = \omega_0{^2} + 2\alpha \theta}\)

\(\mathrm {\omega = 6000\; rpm = 6000(2\pi )/60 = 200\pi \; rad/s \\ \theta = 500\; rev = 500(2\pi) = 1000\pi \; rad \\ 0 = (200\pi)^2 + 2\alpha(1000\pi) \\ \alpha = -20\pi \;rad/s^2}\)

 Which is the appropriate equation to use in this problem? Note that all of the following are correct equations.

(A)    \(\mathrm {\omega = \omega_0 + \alpha t}\)

(B)    \(\mathrm {\omega = \omega_0 + \omega_0t + ½\alpha t^2}\)

(C)    \(\mathrm {\omega^2 = \omega_0{^2} + 2\alpha \theta}\)
 

Either of the equations
\(\mathrm {\omega = \omega_0 + \alpha t}\)
\(\mathrm {\omega = \omega_0 + \omega_0t + ½\alpha t^2}\)

can be used as they both involve t. We will use the first.

In part (a) we calculated that \(\mathrm {\alpha = - 20\pi \; rad/s^2 \;and\; \omega_0 = 200\pi \; rad/s}\)

\(\mathrm {0 = 200\pi  - 20\pi t}\)

\(\mathrm {t = 10 \;s}\)

The moment of inertia is given by

\(\mathrm {I = \sum mr^2}\)

\(\mathrm {I = 1 \times 0.52 + 1 \times 12 + 1 \times 1.52 + 2 \times 22 = 0.25 + 1 + 2.25 + 8 = 11.5 \;kg\; m^2}\)

The moment of inertia is given by

\(\mathrm{I = \sum mr^2}\)

\(\mathrm {I =  3 \times 12 + 1 \times 1.52 + 1 \times 22 = 3 + 2.25 + 4 = 9.25\; kg\; m^2}\)

The rotational equivalent of Newton's 2nd law is

\(\mathrm {\sum \tau = I \alpha}\)

\(\mathrm {For\; a\; uniform\; cylinder\; of\; radius\; r\; and\; mass\; m, I = ½mr^2}\)

\(\mathrm {150\; r = ½mr^2 (15) = ½(12)r^2 (15)}\)

\(\mathrm {r = 10/6 = 1.67 \;m}\)

\(\mathrm {For\; a\; uniform\; cylinder\; of\; mass\; m\; and\; radius\; r, I = ½mr^2}\)

\(\mathrm {In\; part\; (a)\; it\; was\; determined\; that\; r = 1.67\; m}\)

\(\mathrm {Therefore\; I = ½(12)(1.67)^2 = 16.7\; kg\; m^2}\)

Find the total angle turned through in \(\mathrm {1.5 \;min = 90\; s}\)

\(\mathrm {\theta =\theta_0 + \omega_0t + ½\alpha t^2}\)

\(\mathrm {\theta = 0 + 0t + ½(15)(90)^2 = 60750 \;rad}\)

\(\mathrm {Length\; of\; rope = distance\; of\; circumference\; traced\; out = r\theta = 1.67(60750 = 1.01\times 105 \;m)}\)

The moment of inertia of the rod and masses rotated about the centre is \(\mathrm{I = \sum r^2 = 1(1)^2 + 1(1)^2 = 2\; kg\; m^2}\)

\(\mathrm {\omega = 150(2\pi)/60 = 15.7\; rad/s}\)

\(\mathrm {KE = ½I\omega^2 = ½(2)(15.7)^2 = 247\; J}\)

For a thin hoop \(\mathrm {I = mr^2}\)

From conservation of energy

\(\mathrm {PE = KE_{trans} + KE_{rot}}\)

\(\mathrm{mgh = ½mv^2 + ½I\omega^2 = ½mv^2 + ½(mr^2)\omega^2 = ½mv^2 + ½(mr^2)(v/r)^2 = ½mv^2 + ½mv^2 = mv^2}\)

\(\mathrm{Therefore\; gh = v^2}\)

\(\mathrm{v = (gh)^{1/2} = (9.8 \times 15)^{1/2} = 12.1\; m/s}\)

\(\mathrm {2\; rps = 4\pi \; rad/s}\)

\(\mathrm {L = I\omega = I(4\pi) = 14 \\ Therefore\; I = 14/(4\pi) kg\; m^2}\)

\(\mathrm {KE = ½I\omega^2 =  [½(14)/(4\pi)](4\pi)2 = 88\; J}\)

In order to solve part (b) it is useful to have the solution to part (a) handy

I increases by 20% but momentum is conserved

\(\mathrm{I\omega = I'\omega' = 1.2\; I\omega' \\ \omega' = \omega /1.2}\)

\(\mathrm {KE' = ½I'\omega'^2 = ½(1.2 I)(\omega /1.2)^2 = ½I\omega^2(1/1.2)}\)

\(\mathrm {But\; ½I\omega^2 = 88\; J, the\; original\; KE}\)

\(\mathrm{Therefore\; KE' = 88/1.2 = 73 \;}\)

The mass of the wheel is irrelevant.

The change in momentum of the wheel is equal to the product of the moment and the time
i.e., \(\mathrm {\Delta L = \tau t = (100 \times 1) (60) = 6.0\times 10^3 \;kg\; m^2/s}\)

Alternatively if I is the moment of inertia of the wheel then
\(\mathrm {\tau = I \alpha}\)

\(\mathrm {\alpha = (100 \times 1)/I}\)

\(\mathrm {\omega = \omega_0 + \alpha \;t = 0 + (100/I)(60) = 6.0\times 10^3/I}\)

\(\mathrm {L = I\omega = I(6.0\times 10^3)/I = 6.0\times 10^3 \;kg \;m^2/s  \;as \;before}\)