Q: An $8.3\text{ kg}$ mass is attached to a string that has a breaking strength of $1500\text{ N}$. If the mass is whirled in a horizontal circle of radius $80\text{ cm}$, what maximum speed can it have?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: Mass, $m = 8.3\text{ kg}$, $T_{\max} = 1500\text{ N}$,
$r = 80\text{ cm} = 80 \times 10^{-2}\text{ m}$, $g = 9.8\text{ m s}^{-2}$
maximum speed, $v_{\max} = ?$
For horizontal circle,
$F_C = T$
$\frac{m v_{\max}^2}{r} = T_{\max}$
$\frac{8.3 v_{\max}^2}{80 \times 10^{-2}} = 1500$
$v_{\max}^2 = \frac{1500 \times 80 \times 10^{-2}}{8.3}$
$v_{\max} = \sqrt{\frac{1500 \times 80 \times 10^{-2}}{8.3}}$
$= \mathbf{12.02\text{ m s}^{-1}}$
Q: What is the tension in a $1\text{ m}$ string that is spinning $0.5\text{ kg}$ stone in a horizontal circle with $3\text{ rps}$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $m = 0.5\text{ kg}$, $r = 1\text{ m}$, $g = 9.8\text{ m s}^{-2}$, $T = ?$
$\omega = 3\text{ rps } (\text{rev s}^{-1})$
$= 3 \frac{\text{rev}}{\text{s}} = 3 \times \frac{2\pi\text{ rad}}{\text{s}} = 6\pi\text{ rad s}^{-1}$
For horizontal circle,
$F_C = T$
$m r \omega^2 = T$
$T = m r \omega^2 = 0.5 \times 1 \times (6\pi)^2$
$= \mathbf{177.69\text{ N}}$
Q: A $60\text{ cm}$ rope is tied to the handle of a bucket which is then whirled in a vertical circle. The mass of the bucket is $3\text{ kg}$. If the tension of the rope at the lowest point in its path is $50\text{ N}$, find the speed of the bucket at that point.
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $m = 3\text{ kg}$, $g = 9.8\text{ m s}^{-2}$, $r = 60\text{ cm} = 60 \times 10^{-2}\text{ m}$, $v = ?$
Force acting on bucket are:
Tension force, $T = 50\text{ N}$
Weight, $w = mg = 3 \times 9.8 = 29.4\text{ N}$
For vertical circle, at lowest point,
$F_{\text{net}} = T - w = 50 - 29.4 = 20.6\text{ N}$
(Since $F_{\text{net}}$ provides the centripetal acceleration)
$F_C = F_{\text{net}}$
$\frac{m v^2}{r} = F_{\text{net}}$
$\frac{3 v^2}{60 \times 10^{-2}} = 20.6$
$v^2 = \frac{20.6 \times 60 \times 10^{-2}}{3}$
$v = \sqrt{\frac{20.6 \times 60 \times 10^{-2}}{3}}$
$= \mathbf{2.03\text{ m s}^{-1}}$
Q: A $0.15\text{ kg}$ ball on the end of a $1.1\text{ m}$ long cord (negligible mass) is swung in a vertical circle. (i) Determine the minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle. (ii) Calculate the tension in the cord at the bottom of the arc, assuming the ball is moving at twice the speed of part (i).
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $m = 0.15\text{ kg}$, $r = 1.1\text{ m}$, $g = 9.8\text{ m s}^{-2}$
(i) $v = ?$ (ii) $T = ?$
(i) For vertical circle, At the top position,
$F_{\text{net}} = T + w = T + mg$
(Since $F_{\text{net}}$ provides the centripetal acceleration)
At the top, For minimum speed, $T_{\min} = 0$
$F_C = F_{\text{net}}$
$\frac{m v^2}{r} = T + mg$
$\frac{m v_{\min}^2}{r} = mg$
$\frac{v_{\min}^2}{r} = g$
$v_{\min}^2 = g r$
$v_{\min} = \sqrt{g r} = \sqrt{9.8 \times 1.1}$
$= \mathbf{3.283\text{ m s}^{-1}}$
(ii) At the bottom, the speed is twice than at the top.
$v = 2 v_{\min} = 2 \times 3.283 = 6.566\text{ m s}^{-1}$
At the bottom, $F_C = T - w$
$T = F_C + w$
$= \frac{m v^2}{r} + mg$
$= \frac{0.15 \times (6.566)^2}{1.1} + (0.15 \times 9.8)$
$= \mathbf{7.349\text{ N}}$
Q: What is the banking angle for an expressway off-ramp curve of radius $50\text{ m}$ at a limiting speed of $50\text{ km h}^{-1}$?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $r = 50\text{ m}$, $\theta = ?$, $g = 9.8\text{ m s}^{-2}$
$v = 50\text{ km h}^{-1} = 50 \times \frac{10^3\text{ m}}{3600\text{ s}} = 13.89\text{ m s}^{-1}$
$\tan \theta = \frac{v^2}{r g}$
$\tan \theta = \frac{13.89^2}{50 \times 9.8}$
$\theta = \tan^{-1}\left(\frac{13.89^2}{50 \times 9.8}\right)$
$= \mathbf{21.49^\circ \text{ (or) } 21^\circ 29'}$
Q: An athlete weighing $790\text{ N}$ is running a curve at a speed of $6\text{ m s}^{-1}$ in an arc. The radius of curvature of the arc is $5\text{ m}$. Find the centripetal force acting on him.
Which force provides the centripetal force?
What will happen to him if the radius of curvature is smaller?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $w = 790\text{ N}$, $v = 6\text{ m s}^{-1}$, $r = 5\text{ m}$, $g = 9.8\text{ m s}^{-2}$, $F_C = ?$
$m = \frac{w}{g} = \frac{790}{9.8} = 80.61\text{ kg}$
$F_C = \frac{m v^2}{r}$
$= \frac{80.61 \times (6)^2}{5} = \mathbf{580.4\text{ N}}$
⇒ Frictional force between the feet and ground provides the centripetal force.
⇒ If the radius of curvature is smaller, the centripetal force will become greater. He will slip since the frictional force is less than the required centripetal force. [$F_f < F_C$, he will slip]
Q: A jet plane is flying around the airport with a speed of $800\text{ km h}^{-1}$ along a circular path with a radius of $2\text{ km}$. At what angle must the wings of the plane be banked?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $r = 2\text{ km} = 2 \times 10^3\text{ m}$, $\theta = ?$, $g = 9.8\text{ m s}^{-2}$
$v = 800\text{ km h}^{-1} = 800 \times \frac{10^3}{3600} = 222.22\text{ m s}^{-1}$
$\tan \theta = \frac{v^2}{r g}$
$\tan \theta = \frac{222.22^2}{2 \times 10^3 \times 9.8}$
$\theta = \tan^{-1}\left(\frac{222.22^2}{2 \times 10^3 \times 9.8}\right)$
$= \mathbf{68.35^\circ \text{ (or) } 68^\circ 21'}$
Q: An aeroplane is circling above an airport in a horizontal circle at a speed of $400\text{ km h}^{-1}$. The banking angle of the wing is $20^\circ$. What is the radius of the circular path?
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $r = ?$, $\theta = 20^\circ$, $g = 9.8\text{ m s}^{-2}$
$v = 400\text{ km h}^{-1} = 400 \times \frac{10^3}{3600} = 111.11\text{ m s}^{-1}$
$\tan \theta = \frac{v^2}{r g}$
$r = \frac{v^2}{\tan \theta \cdot g} = \frac{111.11^2}{\tan 20^\circ \times 9.8} = \mathbf{3461\text{ m}}$
Q: A curved roadway has a radius of curvature of $200\text{ m}$, and a banking angle of $10^\circ$. What is the highest speed at which a car can round the curve safely? (Neglect the friction between the tire and the road.)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $r = 200\text{ m}$, $\theta = 10^\circ$, $g = 9.8\text{ m s}^{-2}$, $v = ?$
$\tan \theta = \frac{v^2}{r g}$
$v^2 = r g (\tan \theta)$
$v = \sqrt{r g (\tan \theta)}$
$= \sqrt{200 \times 9.8 \times \tan 10^\circ}$
$= \mathbf{18.59\text{ m s}^{-1}}$
Q: In an atom, an electron moves in a circular path around the nucleus. The speed of the electron is approximately $2.2 \times 10^6\text{ m s}^{-1}$. Find the centripetal force acting on the electron as it revolves in a circular orbit of radius $0.53 \times 10^{-10}\text{ m}$. ($m_e = 9.1 \times 10^{-31}\text{ kg}$)
Ans: (အဖြေနှင့် တွက်နည်းကြည့်ရန် နှိပ်ပါ)
Ans: $v = 2.2 \times 10^6\text{ m s}^{-1}$, $m_e = 9.1 \times 10^{-31}\text{ kg}$
$r = 0.53 \times 10^{-10}\text{ m}$, $F_C = ?$
$F_C = \frac{m v^2}{r}$
$= \frac{9.1 \times 10^{-31} \times (2.2 \times 10^6)^2}{0.53 \times 10^{-10}}$
$= \mathbf{8.3101 \times 10^{-8}\text{ N}}$