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Q451
Allen
Dynamics of Circular Motion
Numerical
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Correct Answer: $T = 6.6 N, v_{max} = 35 ms^{-1}$
Explanation:
Ans. $T = 6.6N, v_{\mathrm{max}} = 35ms^{-1}$
$v = r \omega = (1.5) (2 \pi) \left(\frac {4 0}{6 0}\right)$
$T = \frac {m v ^ {2}}{R} = \frac {0.25 \times 4 \pi^ {2}}{1.5} = 6.6 N$
$2 0 0 = \frac {m v _ {m a x} ^ {2}}{R}$
$v _ {\max} = 3 5 m s ^ {- 1}$
$v = r \omega = (1.5) (2 \pi) \left(\frac {4 0}{6 0}\right)$
$T = \frac {m v ^ {2}}{R} = \frac {0.25 \times 4 \pi^ {2}}{1.5} = 6.6 N$
$2 0 0 = \frac {m v _ {m a x} ^ {2}}{R}$
$v _ {\max} = 3 5 m s ^ {- 1}$
Q452
Allen
Dynamics of Circular Motion
Numerical
A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed. Consider a particle of mass 1g sticking at the outer end of a blade. What is the net force on it, when the fan runs at full speed? Who exerts this force on the particle? How much force does the particle exert on the blade in the plane of motion?
Correct Answer: $\frac{15\pi^{2}}{10}=14.8N$ , This force is exerted by blade of fan and equal force is exerted by particle on blade in same magnitude but opposite in direction.
Explanation:
Ans. $\frac{15\pi^{2}}{10}=14.8N$ , This force is exerted by blade of fan and equal force is exerted by particle on blade in same magnitude but opposite in direction. $\omega=2\pi n=\frac{2\pi\times1500}{60}$ rad/sec $r=\frac{d}{2}=60\ cm=0.6\ m$ $m=1\ g=10^{-3}\ kg$ $F=m\omega^{2}r=10^{-3}\times\left(\frac{2\pi\times1500}{60}\right)^{2}\times0.6=\frac{15\pi^{2}}{10}=14.8$
This force is exerted by blade of fan and equal force is exerted by particle on blade in same magnitude but opposite in direction.
This force is exerted by blade of fan and equal force is exerted by particle on blade in same magnitude but opposite in direction.
Q453
Allen
Dynamics of Circular Motion
Numerical
A table with smooth horizontal surface is placed in a cabin which moves in a circle of a large radius R (figure). A smooth pulley of small radius is fastened to the table. Two masses m and 2m placed on the table are connected through a string going over the pulley. Initially the masses are held by a person with the strings along the outward radius and then the system is released from rest (with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension in the string.
Correct Answer: $\frac{4m\omega^{2}R}{3}$
Explanation:
Ans.
$T - m \omega^ {2} R = m a....(i)$
$2 m \omega^ {2} R - T = 2 m a$
(i) + (ii)
$m w ^ {2} R = 3 m a....(ii)$
$a = \frac {\omega^ {2} R}{3}$
Put in eq (i)
$T - m \omega^ {2} R = a = \frac {\omega^ {2} R}{3}$
$T = m \omega^ {2} R + \frac {m \omega^ {2} R}{3} = \frac {4 m \omega^ {2} R}{3}$
$T - m \omega^ {2} R = m a....(i)$
$2 m \omega^ {2} R - T = 2 m a$
(i) + (ii)
$m w ^ {2} R = 3 m a....(ii)$
$a = \frac {\omega^ {2} R}{3}$
Put in eq (i)
$T - m \omega^ {2} R = a = \frac {\omega^ {2} R}{3}$
$T = m \omega^ {2} R + \frac {m \omega^ {2} R}{3} = \frac {4 m \omega^ {2} R}{3}$
Q454
Allen
Dynamics of Circular Motion
Numerical
A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The co-efficient of static-friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?
Correct Answer: Yes, $a_{c}=8$ , $\mu g=1$
Explanation:
Ans. Yes, $a_{c}=8$ , $\mu g=1$ $v=\sqrt{\mu rg}$ So, required value of $\mu=\frac{v^{2}}{rg}=\frac{25}{3\times10}=\frac{5}{6}=0.8$ $\mu=0.1$ So, cyclist will slip.
Q455
Allen
Dynamics of Circular Motion
Numerical
A cyclist moving with a speed of 4.9 m/s on a level road can take a sharp circular turn of radius 4 m, then find the minimum value of coefficient of friction between the cycle tyres and road. $(g = 9.8 \, m/s^{2})$
Correct Answer: $\frac{49}{80}$
Explanation:
Ans. $\frac{49}{80}$ Coefficient of friction $\mu \geq \frac{v^2}{rg}$ $\Rightarrow \mu_{\min} = \frac{(4.9)^2}{4\times 9.8} = \frac{49}{80}$
Q456
Allen
Dynamics of Circular Motion
Numerical
When the road is dry and coefficient of friction is $\mu$ , the maximum speed of a car in a circular path is $10 \, ms^{-1}$ . If the road becomes wet and coefficient of friction become $\frac{\mu}{2}$ , what is the maximum speed permitted?
Correct Answer: $5\sqrt{2} ms^{-1}$
Explanation:
Ans. $5\sqrt{2} ms^{-1}$ For safe driving, $v_{\max} = \sqrt{\mu rg}$ $10 = \sqrt{\mu rg}$ For wet road $v' = \sqrt{\frac{\mu}{2} rg} = \frac{10}{\sqrt{2}} = 5\sqrt{2} m / s$
Q457
Allen
Dynamics of Circular Motion
Numerical
A block of mass m is sitting on a rotating smooth wedge. The wedge rotates with constant angular velocity $\omega$ around the axis shown in figure. Calculate the value of w (in rad/s) such that the block stays at constant height h = 2.5 m.
Correct Answer: 2.0
Explanation:
Ans. 2.0
FBD of block in ground frame
$N\cos \theta = mg$
$\begin{array}{rl} & N\sin \theta = m\omega^2 r\quad [\mathrm{centripetalforce}]\\ \Rightarrow & \tan \theta = \omega^2 r / g\\ \Rightarrow & \omega = \sqrt{\frac{g\tan\theta}{r}} = \sqrt{\frac{g}{h}} (\tan \theta) = 2\times 1 = 2\mathrm{rad / s} \end{array}$
FBD of block in ground frame
$N\cos \theta = mg$
$\begin{array}{rl} & N\sin \theta = m\omega^2 r\quad [\mathrm{centripetalforce}]\\ \Rightarrow & \tan \theta = \omega^2 r / g\\ \Rightarrow & \omega = \sqrt{\frac{g\tan\theta}{r}} = \sqrt{\frac{g}{h}} (\tan \theta) = 2\times 1 = 2\mathrm{rad / s} \end{array}$
Q458
Allen
Dynamics of Circular Motion
Numerical
For traffic moving at 60 km/hr, if the radius of the curve is 0.1 km, what is the correct angle of banking of the road? $(g = 10 \, m/s^{2})$
Correct Answer: $\theta = tan^{-1}\left(\frac{5}{18}\right)$
Explanation:
Ans. $\theta = tan^{-1}\left(\frac{5}{18}\right)$
In case of banking $\tan \theta = \frac{v^2}{rg}$ Here $v = 60km / hr = 60\times \frac{5}{18} ms - 1$ $= \frac{50}{3} ms^{-1}r = 0.1km = 100m$ So $\tan \theta = \frac{\left(\frac{50}{3}\right)^2}{100\times 10} = \frac{5}{18}\quad \Rightarrow \quad \theta = \tan^{-1}\left(\frac{5}{18}\right)$
In case of banking $\tan \theta = \frac{v^2}{rg}$ Here $v = 60km / hr = 60\times \frac{5}{18} ms - 1$ $= \frac{50}{3} ms^{-1}r = 0.1km = 100m$ So $\tan \theta = \frac{\left(\frac{50}{3}\right)^2}{100\times 10} = \frac{5}{18}\quad \Rightarrow \quad \theta = \tan^{-1}\left(\frac{5}{18}\right)$
Q459
Allen
Dynamics of Circular Motion
Numerical
When an airplane is driven forward by its engines, the air flowing over its wings generates an upward force called lift. When this lift balances the weight of the plane, it can fly horizontally. A plane banks, that is, tilts its wings to the side, to turn, thus using the lift force another way. Note that it must then increase speed to generate more lift if it wants to stay at the same altitude while turning.
A jet airliner remains at constant altitude at a constant speed of v = 720 km/hr while banking at with respect to the horizon to turn in a circle. The radius (in km) of the circle in which the plane moves is given by R. Find the value of $\sqrt{R}$ .
Correct Answer: 4
Explanation:
Ans. 4
$F\cos \theta = mg$ ... (i)
and $F\sin \theta = \frac{mv^2}{R}$ ...(ii)
By solving equation (i) and (ii)
$R = 1 6 \mathrm{km}$
$F\cos \theta = mg$ ... (i)
and $F\sin \theta = \frac{mv^2}{R}$ ...(ii)
By solving equation (i) and (ii)
$R = 1 6 \mathrm{km}$
Q460
Allen
Dynamics of Circular Motion
Numerical
A mosquito is sitting on an L.P. record of a gramophone disc rotating on a turn table at $33\frac{1}{3}$ revolution per minute. The distance of the mosquito from the center of the disc is 10 cm. Find minimum friction coefficient between the record and the mosquito.
Take $g = 10 \, m/s^{2}$ .
Correct Answer: $\frac{\pi^2}{81}$
Explanation:
Ans. $\frac{\pi^2}{81}$
$n = 3 3 \frac {1}{3} r e v / m i n$
$\omega = \frac {2 \pi n}{6 0} = \frac {2 \pi \times 3 3 \frac {1}{3}}{6 0} \mathrm{rad/sec.,}$
r = 1 0 c m = 0.1 m.
For no slipping,
$f \geq m \omega^ {2} r$
$\mu m g \geq m \omega^ {2} r$
$\mu \geq \frac {\omega^ {2} r}{g}$
$\mu \geq \left(\frac {2 \pi 1 0 0}{6 0 \times 3}\right) ^ {2} \frac {0.1}{1 0}$
$\mu \geq \frac {\pi^ {2}}{8 1}$
$n = 3 3 \frac {1}{3} r e v / m i n$
$\omega = \frac {2 \pi n}{6 0} = \frac {2 \pi \times 3 3 \frac {1}{3}}{6 0} \mathrm{rad/sec.,}$
r = 1 0 c m = 0.1 m.
For no slipping,
$f \geq m \omega^ {2} r$
$\mu m g \geq m \omega^ {2} r$
$\mu \geq \frac {\omega^ {2} r}{g}$
$\mu \geq \left(\frac {2 \pi 1 0 0}{6 0 \times 3}\right) ^ {2} \frac {0.1}{1 0}$
$\mu \geq \frac {\pi^ {2}}{8 1}$
Q461
Allen
Dynamics of Circular Motion
Numerical
A motorcyclist wants to drive on the vertical surface of wooden āwellā of radius 5 m, in horizontal plane with speed of $5\sqrt{5}$ m/s. Find the minimum value of coefficient of friction between the tyres and the wall of the well. (take $g = 10 \, m/s^{2}$ )
Correct Answer: $\frac{2}{5}$
Explanation:
Ans. $\frac{2}{5}$
$N = \frac {m v ^ {2}}{r}$
Given $r = 5 \, m$ , $v = 5\sqrt{5} \, m/s$
for no slipping,
$f \geq m g$
$\mu_ {\mathrm{min}} N = m g$
$\mu_ {\mathrm{min}} = \frac {m g}{N} = \frac {r g}{v ^ {2}}$
$\mu_ {\mathrm{min}} = \frac {5 \times 1 0}{(5 \sqrt {5}) ^ {2}} = \frac {2}{5}$
$N = \frac {m v ^ {2}}{r}$
Given $r = 5 \, m$ , $v = 5\sqrt{5} \, m/s$
for no slipping,
$f \geq m g$
$\mu_ {\mathrm{min}} N = m g$
$\mu_ {\mathrm{min}} = \frac {m g}{N} = \frac {r g}{v ^ {2}}$
$\mu_ {\mathrm{min}} = \frac {5 \times 1 0}{(5 \sqrt {5}) ^ {2}} = \frac {2}{5}$
Q462
Allen
Dynamics of Circular Motion
Numerical
A train has to negotiate a curve of radius 400 m. By how much height should the outer rail be raised with respect to inner rail for a speed of 48 km/hr? The distance between the rails is 1 m :
Correct Answer: $\frac{2}{45} m$
Explanation:
Ans. $\frac{2}{45} m$ $v = 48 km/hr = 40/3 m/s$ For safe turn without friction $\tan \theta = \frac{v^2}{rg} = \frac{h}{x}$ Given $x \approx \ell = 1m$ $\Rightarrow h = \frac{v^2}{rg} = \frac{(40/3)^2}{400 \times 10} = \frac{2}{45} m$
Q463
Allen
Dynamics of Circular Motion
Numerical
A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?
Between $\sqrt{\frac{50}{3}}\times\frac{18}{5}=14.7$ km/h and 54 km/hr
Correct Answer: $\sqrt{\frac{50}{3}}\times \frac{18}{5} = 14.7km / h$
Explanation:
Ans. $\sqrt{\frac{50}{3}} \times \frac{18}{5} = 14.7 \, \text{km/h}$
$v = 3 6 \mathrm{km} / h r = 1 0 \mathrm{m} / \mathrm{s}$
$\tan \theta = \frac {v ^ {2}}{R g} = \frac {1 0 \times 1 0}{2 0 \times 1 0} = \frac {1}{2}$
$v _ {\max} = \sqrt {R g \left(\frac {\tan \theta + \mu}{1 - \mu \tan \theta}\right)} = 1 5 m / s = 5 4 k m / h r \quad \text { Ans. }$
$\begin{array}{r l} v _ {\min} & = \sqrt {R g \left(\frac {\tan \theta - \mu}{1 + \mu \tan \theta}\right)} = \sqrt {\frac {2 0 \times 1 0 (0.5 - 0.4)}{(1 + 0.5 \times 0.4)}} \\ & = \sqrt {\frac {2 0 \times 1 0 \times 0.1}{1.2}} = \sqrt {\frac {5 0}{3}} \times \frac {1 8}{5} k m / h r \mathbf {A n s}. \end{array}$
$v = 3 6 \mathrm{km} / h r = 1 0 \mathrm{m} / \mathrm{s}$
$\tan \theta = \frac {v ^ {2}}{R g} = \frac {1 0 \times 1 0}{2 0 \times 1 0} = \frac {1}{2}$
$v _ {\max} = \sqrt {R g \left(\frac {\tan \theta + \mu}{1 - \mu \tan \theta}\right)} = 1 5 m / s = 5 4 k m / h r \quad \text { Ans. }$
$\begin{array}{r l} v _ {\min} & = \sqrt {R g \left(\frac {\tan \theta - \mu}{1 + \mu \tan \theta}\right)} = \sqrt {\frac {2 0 \times 1 0 (0.5 - 0.4)}{(1 + 0.5 \times 0.4)}} \\ & = \sqrt {\frac {2 0 \times 1 0 \times 0.1}{1.2}} = \sqrt {\frac {5 0}{3}} \times \frac {1 8}{5} k m / h r \mathbf {A n s}. \end{array}$
Q464
Allen
Dynamics of Circular Motion
Numerical
A circular racetrack of radius 300 m is banked at an angle of $15^{\circ}$ . If the coefficient of friction between the wheels of a race car and the road is 0.2.
(A) What is the optimum speed of the race car to avoid wear and tear on its tires?
(B) What is the maximum permissible speed to avoid slipping?
Correct Answer: (a) $28m / s$ (b) $38.19m / s$
Explanation:
Ans. (a) 28 m/s (b) 38.19 m/s
a) Optimum speed of car is $\tan\theta=\frac{v^{2}}{rg}$ $v^{2}=rg\tan\theta$ $=10\times300\times\tan15$ $v^{2}=780\ m/s$ v=28 m/s
b) Maximum permissible speed to avoid slipping is
$\begin{array}{r l} v _ {m} & = \sqrt {R g \left(\frac {\mu + \tan \theta}{1 - \mu \tan \theta}\right)} \\ & = \sqrt {3 0 0 \times 1 0 \left(\frac {0 . 2 + \tan 1 5}{1 - 0 . 2 \tan 1 5}\right)} \\ & = \sqrt {1 4 5 8 . 7} \\ & = 3 8. 1 9 m / s \end{array}$
a) Optimum speed of car is $\tan\theta=\frac{v^{2}}{rg}$ $v^{2}=rg\tan\theta$ $=10\times300\times\tan15$ $v^{2}=780\ m/s$ v=28 m/s
b) Maximum permissible speed to avoid slipping is
$\begin{array}{r l} v _ {m} & = \sqrt {R g \left(\frac {\mu + \tan \theta}{1 - \mu \tan \theta}\right)} \\ & = \sqrt {3 0 0 \times 1 0 \left(\frac {0 . 2 + \tan 1 5}{1 - 0 . 2 \tan 1 5}\right)} \\ & = \sqrt {1 4 5 8 . 7} \\ & = 3 8. 1 9 m / s \end{array}$
Q465
Allen
Dynamics of Circular Motion
Numerical
A block of mass m = 20 kg is kept at a distance R = 1m from central axis of rotation of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration, $\alpha = 3$ rad/sec $^{2}$ . The friction coefficient between block and table is $\mu = 0.5$ . At time $t = \frac{x}{3}$ sec from starting of motion (i.e. t = 0 sec) the block is just about to slip. Find the value of x.
Correct Answer: 2
Explanation:
Ans. 2
$\mu m g = m \sqrt {a _ {t} ^ {2} + a _ {n} ^ {2}}$
$r \omega^ {2} = (\alpha t) ^ {2} = x ^ {2}$
$(0.5 \times 1 0) ^ {2} = x ^ {4} + 9$
$x ^ {4} = 1 6$
x = 2
$\mu m g = m \sqrt {a _ {t} ^ {2} + a _ {n} ^ {2}}$
$r \omega^ {2} = (\alpha t) ^ {2} = x ^ {2}$
$(0.5 \times 1 0) ^ {2} = x ^ {4} + 9$
$x ^ {4} = 1 6$
x = 2
Q466
Allen
Dynamics of Circular Motion
Numerical
A car goes on a horizontal circular road of radius R, the speed increasing at constant rate $\frac{dv}{dt}=a$ . The friction coefficient between the road and the tyre is $\mu$ . Find the speed at which the car will skid.
Correct Answer: $V = \left[\left(\mu^2 g^2 - a^2\right)R^2\right]^{\frac{1}{4}}$
Explanation:
Ans. $V = \left[\left(\mu^2 g^2 - a^2\right) R^2\right]^{\frac{1}{4}}$
$a _ {R} = \sqrt {a _ {N} ^ {2} + a _ {t} ^ {2}}$
$a _ {N} = \frac {V ^ {2}}{R}, a _ {t} = \frac {d V}{d t} = a$
Friction, $f = ma_{R}$
$\mu N = m \sqrt {\left(\frac {V ^ {2}}{R}\right) ^ {2} + \left(\frac {d V}{d t}\right) ^ {2}}$
$\mu N = m \sqrt {\frac {V ^ {4}}{R ^ {2}} + a ^ {2}}$
$V = \left[ \left(\mu^ {2} g ^ {2} - a ^ {2}\right) R ^ {2} \right] ^ {\frac {1}{4}}$
$a _ {R} = \sqrt {a _ {N} ^ {2} + a _ {t} ^ {2}}$
$a _ {N} = \frac {V ^ {2}}{R}, a _ {t} = \frac {d V}{d t} = a$
Friction, $f = ma_{R}$
$\mu N = m \sqrt {\left(\frac {V ^ {2}}{R}\right) ^ {2} + \left(\frac {d V}{d t}\right) ^ {2}}$
$\mu N = m \sqrt {\frac {V ^ {4}}{R ^ {2}} + a ^ {2}}$
$V = \left[ \left(\mu^ {2} g ^ {2} - a ^ {2}\right) R ^ {2} \right] ^ {\frac {1}{4}}$
Q467
Allen
Dynamics of Circular Motion
Numerical
A motorcycle has to move with a constant speed on an overbridge which is in the form of a circular are of radius R and has a total length L. Suppose the motorcycle starts from the highest point.
(A) What can its maximum velocity be for which the contact with the road is not broken at the highest point?
(B) If the motorcycle goes at speed $1/\sqrt{2}$ times the maximum found in part a), where will it lose the contact with the road?
(C) What maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge?
Correct Answer: (a) $v = \sqrt{Rg}$ , (b) $\frac{\pi R}{3}$ , (c) $V = \sqrt{gR\cos\left(\frac{L}{2R}\right)}$
Explanation:
Ans. a) $v = \sqrt{Rg}$ , b) $\frac{\pi R}{3}$ , c) $V = \sqrt{gR\cos\left(\frac{L}{2R}\right)}$
a) At highest point
$m g = \frac {m v ^ {2}}{R}$
$\Rightarrow v = \sqrt {R g}$
b) $V = \frac{1}{\sqrt{2}} \sqrt{Rg}$
Losses contact at angle $\theta$
$m g \cos \theta = \frac {m V ^ {2}}{R}$
c) $\alpha=\frac{L}{2R}$
So, $\frac{mV^2}{R} = mg\cos \theta$
$V = \sqrt {g R \cos \left(\frac {L}{2 R}\right)}$
a) At highest point
$m g = \frac {m v ^ {2}}{R}$
$\Rightarrow v = \sqrt {R g}$
b) $V = \frac{1}{\sqrt{2}} \sqrt{Rg}$
Losses contact at angle $\theta$
$m g \cos \theta = \frac {m V ^ {2}}{R}$
c) $\alpha=\frac{L}{2R}$
So, $\frac{mV^2}{R} = mg\cos \theta$
$V = \sqrt {g R \cos \left(\frac {L}{2 R}\right)}$
Q468
Allen
Dynamics of Circular Motion
Numerical
A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is $\mu$ . The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration $\alpha$ , at what angular speed will the block slip?
Correct Answer: (A) $\omega_{\max} = \sqrt{\frac{\mu g}{L}}$ , (B) $\omega = \left(\left(\frac{\mu g}{L}\right)^2 - \alpha^2\right)^{\frac{1}{4}}$
Explanation:
Ans. (A) $\omega_{\max} = \sqrt{\frac{\mu g}{L}}$ , B) $\omega = \left(\left(\frac{\mu g}{L}\right)^2 - \alpha^2\right)^{\frac{1}{4}}$
A)
$f _ {\max} = \mu N = (\mu m g)$
$f = (m \omega^ {2} L)$
$f \leq f _ {\mathrm{max}}$
$m \omega^ {2} L \leq \mu m g$
$\omega^ {2} \leq \frac {\mu g}{L} \Rightarrow \omega \leq \left(\sqrt {\frac {\mu g}{L}}\right)$
$\omega_ {\mathrm{max}} = \sqrt {\frac {\mu g}{L}}$
B)
$a _ {n e t} = \sqrt {a _ {C} ^ {2} + a _ {t} ^ {2}} = \sqrt {\left(\omega^ {2} L\right) ^ {2} + (\alpha L) ^ {2}}$
$f = m a _ {n e t} = m \sqrt {\left(\omega^ {2} L\right) ^ {2} + (\alpha L) ^ {2}}$
$f \leq f _ {\mathrm{max}}$
$f = f _ {\mathrm{max}}$
$m \sqrt {\left(\omega^ {2} L\right) ^ {2} + \left(\alpha L\right) ^ {2}} = \mu m g$
$\omega \left(\left(\frac {\mu g}{L}\right) ^ {2} - \alpha^ {2}\right) ^ {\frac {1}{4}}$
A)
$f _ {\max} = \mu N = (\mu m g)$
$f = (m \omega^ {2} L)$
$f \leq f _ {\mathrm{max}}$
$m \omega^ {2} L \leq \mu m g$
$\omega^ {2} \leq \frac {\mu g}{L} \Rightarrow \omega \leq \left(\sqrt {\frac {\mu g}{L}}\right)$
$\omega_ {\mathrm{max}} = \sqrt {\frac {\mu g}{L}}$
B)
$a _ {n e t} = \sqrt {a _ {C} ^ {2} + a _ {t} ^ {2}} = \sqrt {\left(\omega^ {2} L\right) ^ {2} + (\alpha L) ^ {2}}$
$f = m a _ {n e t} = m \sqrt {\left(\omega^ {2} L\right) ^ {2} + (\alpha L) ^ {2}}$
$f \leq f _ {\mathrm{max}}$
$f = f _ {\mathrm{max}}$
$m \sqrt {\left(\omega^ {2} L\right) ^ {2} + \left(\alpha L\right) ^ {2}} = \mu m g$
$\omega \left(\left(\frac {\mu g}{L}\right) ^ {2} - \alpha^ {2}\right) ^ {\frac {1}{4}}$
Q469
Allen
Dynamics of Circular Motion
Numerical
A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure. Each part subtends a rights angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the track.
(A) Find the normal contact force by the road on the cycle when it is at B and at D.
(B) Find the force of friction exerted by the track on the tyres when the cycle is at B, C, and D.
(C) Find the normal force between the cycle crosses C.
(D) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed? (Take $g = 10 \, m/s^2$ )
(A) Find the normal contact force by the road on the cycle when it is at B and at D.
(B) Find the force of friction exerted by the track on the tyres when the cycle is at B, C, and D.
(C) Find the normal force between the cycle crosses C.
(D) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed? (Take $g = 10 \, m/s^2$ )
Correct Answer: (a) 975 N, 1025 N (b) 707 N (c) 683 N (d) $\mu = 1.037$
Explanation:
Ans. (a) 975 N, 1025 N (b) 707 N (c) 683 N (d) $\mu = 1.037$
a) At $B, N = mg - \frac{mv^2}{R}$ At $D, N = mg + \frac{mv^2}{R}$
b) Friction force is zero (no sliding tendency)
At C, mg sin Īø = F
c) Just before, $mg\cos \theta - N = \frac{mv^2}{R}$
d) Minimum coefficient of friction $\Rightarrow \mu N = mg\sin \theta$
a) At $B, N = mg - \frac{mv^2}{R}$ At $D, N = mg + \frac{mv^2}{R}$
b) Friction force is zero (no sliding tendency)
At C, mg sin Īø = F
c) Just before, $mg\cos \theta - N = \frac{mv^2}{R}$
d) Minimum coefficient of friction $\Rightarrow \mu N = mg\sin \theta$
Q470
Allen
Dynamics of Circular Motion
Numerical
A table with smooth horizontal surface is turning at an angular speed $\omega$ about its axis. A groove is made on the surface along a radius and a particle is gently placed inside the groove at a distance a from the centre. Find the speed of the particle as its distance from the centre becomes L.
Correct Answer: $v = \omega \sqrt{L^2 - a^2}$
Explanation:
Ans. $v = \omega \sqrt{L^{2} - a^{2}}$
The situation is shown in figure.
Thus, the acceleration along the X-axis is
$a = \frac {F}{m} = \frac {m \omega^ {2} x}{m} = \omega^ {2} x$
or $\frac{dv}{dt} = \omega^2 x$
or $\frac{dv}{dx}\cdot \frac{dx}{dt} = \omega^2 x$
or $\frac{dv}{dx}\cdot v = \omega^2 x$
or $v dv = \omega^2 x dx$
or $\int_0^v v dv = \int_a^L\omega^2 x dx$
or $\left[\frac{1}{2}v^{2}\right]_{0}^{v}=\left[\frac{1}{2}\omega^{2}x^{2}\right]_{a}^{L}$
or $\frac{v^2}{2} = \frac{1}{2}\omega^2 (L^2 -a^2)$
or $v = \omega \sqrt{L^2 - a^2}$
The situation is shown in figure.
Thus, the acceleration along the X-axis is
$a = \frac {F}{m} = \frac {m \omega^ {2} x}{m} = \omega^ {2} x$
or $\frac{dv}{dt} = \omega^2 x$
or $\frac{dv}{dx}\cdot \frac{dx}{dt} = \omega^2 x$
or $\frac{dv}{dx}\cdot v = \omega^2 x$
or $v dv = \omega^2 x dx$
or $\int_0^v v dv = \int_a^L\omega^2 x dx$
or $\left[\frac{1}{2}v^{2}\right]_{0}^{v}=\left[\frac{1}{2}\omega^{2}x^{2}\right]_{a}^{L}$
or $\frac{v^2}{2} = \frac{1}{2}\omega^2 (L^2 -a^2)$
or $v = \omega \sqrt{L^2 - a^2}$
Q471
Allen
Work
Numerical
A sleeve of mass 2 kg at origin can move on wire of parabolic shape $x^{2} = 4y$ . Two forces $F_{1} = 6N$ and $F_{2} = 8N$ are applied on the sleeve. $F_{1}$ is constant and is in x-direction. $F_{2}$ is constant in direction and magnitude. Body is displaced from origin to x = 4, then net work done by $F_{1}$ and $F_{2}$ is
Correct Answer: 67.7 J
Explanation:
Ans. 67.7 J
$W _ {1} = \vec {F} _ {1}. \vec {d} _ {1} = (6) (4) = 2 4 N$
$W_{2} = \vec{F}_{2}.\vec{d}_{2}$ where $= 8\cos 30i + 8\sin 30j$ $= (4\sqrt{3}\hat{i} +4\hat{j}),\vec{s} = (4\hat{i} +4\hat{j})$
$\Rightarrow W_{2} = 43.7J$ therefore $W_{\mathrm{net}} = 67.7J$
$W _ {1} = \vec {F} _ {1}. \vec {d} _ {1} = (6) (4) = 2 4 N$
$W_{2} = \vec{F}_{2}.\vec{d}_{2}$ where $= 8\cos 30i + 8\sin 30j$ $= (4\sqrt{3}\hat{i} +4\hat{j}),\vec{s} = (4\hat{i} +4\hat{j})$
$\Rightarrow W_{2} = 43.7J$ therefore $W_{\mathrm{net}} = 67.7J$
Q472
Allen
Work
Numerical
A particle is subject to a force $F_{x}$ that varies with position as in figure. Find the work done by the force on the body as it moves (a) from x = 0 to x = 5.00 m, (b) from x = 5.00 m to x = 10.0 m, and (c) from x = 10.0 m to x = 15.0 m. (d) What is the total work done by the force over the distance x = 0 to x = 15.0 m?
Correct Answer: (a) 7.5 J; (b) 15 J; (c) 7.5 J; (d) 30 J
Explanation:
Ans. (a) 7.5 J; (b) 15 J; (c) 7.5 J; (d) 30 J
W = area under the graph.
W = area under the graph.
Q473
Allen
Work
Numerical
A spring, which is initially in its unstretched condition, is first stretched by a length x and then again by a further length x. The work done in the first case is $W_{1}$ and in the second case is $W_{2}$ . Find $\frac{W_{2}}{W_{1}}$ .
Correct Answer: 3
Explanation:
Ans. 3
$W _ {1} = \frac {1}{2} k x ^ {2}$
$W_{2} = \frac{1}{2} k\left[4x^{2} - x^{2}\right] = \frac{3}{2} kx^{2}$
$W _ {1} = \frac {1}{2} k x ^ {2}$
$W_{2} = \frac{1}{2} k\left[4x^{2} - x^{2}\right] = \frac{3}{2} kx^{2}$
Q474
Allen
Work Energy Theorem
Numerical
Position-time graph of a particle of mass 2 kg is shown in figure. Total work done on the particle from t = 0 to t = 4s is
Correct Answer: 0
Explanation:
Ans. 0
Total work done = change in kinetic energy
$= \frac {1}{2} m v _ {2} ^ {2} - \frac {1}{2} m v _ {1} ^ {2} = \frac {1}{2} (2) \times (1) ^ {2} - \frac {1}{2} \times 2 \times (1) ^ {2} = 0$
Total work done = change in kinetic energy
$= \frac {1}{2} m v _ {2} ^ {2} - \frac {1}{2} m v _ {1} ^ {2} = \frac {1}{2} (2) \times (1) ^ {2} - \frac {1}{2} \times 2 \times (1) ^ {2} = 0$
Q475
Allen
Work Energy Theorem
Numerical
A mass m slides from rest at height h down a smooth curved surface which becomes horizontal at zero height (see figure). A spring is fixed horizontally on the level part of the surface. The spring constant is k N/m. When the mass encounters the spring it compresses it by an amount x = h/10. If m = 1 kg, h = 5m then find k /100.
Correct Answer: 4
Explanation:
Ans. 4
$m g h = \frac {1}{2} k x ^ {2}$
$1 \times 10 \times 5 = \frac{1}{2} \times k \times \frac{1}{4}$
$m g h = \frac {1}{2} k x ^ {2}$
$1 \times 10 \times 5 = \frac{1}{2} \times k \times \frac{1}{4}$