Circular Kinematics(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

CIRCULAR KINEMATICS MCQs

Total Questions : 30 | Page 1 of 3 pages

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/ $s^{2}$ ) $^i$ + (4.00 m/ $s^{2}$ ) $^j$ . At that instant and in unit-vector notation, what is the acceleration of the wallet? IIT JEE- 2001

2 $^i$ + 4 $^j$
4 $^i$ + 2 $^j$
3 $^i$ + 6 $^j$
3 $\sqrt{5}$ $^i$ + 3 $\sqrt{5}$ $^j$

Discuss Question

Answer: Option C. -> 3

^i

+ 6

^j

:
C

$\frac{d θ}{d t}$ is constant.

In other words in uniform circular motion the angular velocity remains constant body doesn't have any tangential acceleration but normal acceleartion.

a_{N} = \frac{v^{2}}{R} o r ω^{2} R

F o r p u r s e a_{N} = \sqrt{(2)^{2} + (4)^{2}} = \sqrt{20}; R = 2

\Rightarrow \sqrt{20} = ω^{2} 2

\Rightarrow ω^{2} = \sqrt{5}

F o r w a l l e t a_{N} = ω^{2} R

Hence

ω

is same
But~R=3

\Rightarrow a_{N} = \sqrt{5} \times 3

a_{N} = 3 \sqrt{5}

So the above answer matches with the magnitude of third option in the given answers.

Question 2.

A balloon starts rising from the surface of the Earth. The ascension rate is constant and equal to v_o. Due to the wind the balloon gathers the horizontal velocity component v_x=ay, where a is a constant and y is the height of ascent. Find the total, tangential, and normal accelerations of the balloon.

$a_{r} = a v_{0}; a_{t} = \frac{a v_{0}}{\sqrt{1 + {(\frac{a y}{v_{0}})}^{2}}}; a_{N} = a v_{0}$
$a_{r} = a v_{0}; a_{t} = a v_{0}; a_{N} = 0$
$a_{r} = a v_{0}; a_{t} = \frac{a^{2} y}{\sqrt{1 + \frac{a^{2} y^{2}}{v_{0}^{2}}}}; a_{N} = \frac{a v_{0}}{\sqrt{(1 + \frac{a^{2} y^{2}}{v_{0}^{2}})}}$
$a_{r} = \frac{a^{2} y}{\sqrt{1} + \frac{a^{2} y^{2}}{v_{0}^{2}}}; a_{t} = a v_{0}; a_{N} = a v_{0}$

Discuss Question

Answer: Option C. ->

a_{r} = a v_{0}; a_{t} = \frac{a^{2} y}{\sqrt{1 + \frac{a^{2} y^{2}}{v_{0}^{2}}}}; a_{N} = \frac{a v_{0}}{\sqrt{(1 + \frac{a^{2} y^{2}}{v_{0}^{2}})}}

:
C

The path of the balloon will look something like this

After t sec ballon would have gone a height of $v_{0} t$ then at that very instance the balloon's $v_{x}$ will be
$v_{x} = a v_{0} t \Rightarrow a_{x} = a v_{0}$
$V_{y} = v_{0}$
$a_{y} = 0$
$a_{r} = t o t a l a c c e l e r a t i o n = a v_{0} . . . . . . . (1)$
$v_{t a n g e n t i c a l}$ is actually the resultant velocity
$\Rightarrow v_{t} = \sqrt{v^{2} 0 + (a v_{0} t)^{2}}$
$\Rightarrow v_{t} = v_{0} \sqrt{1 + a^{2} t^{2}}$
$a_{t} = \frac{d v_{t}}{d t} = \frac{v_{0} a^{2} t}{\sqrt{1 + a^{2} t^{2}}} \Rightarrow \frac{a^{2} y}{\sqrt{1 + {(\frac{a y}{v_{0}})}^{2}}}$
Now we know $a_{t}^{2} + a_{N}^{2} = a_{r}^{2}$
$\Rightarrow \frac{v_{0}^{2} a^{4} t^{2}}{(1 + a^{2} t^{2})} + a_{N}^{2} = a^{2} v_{0}^{2}$
$a_{N} = \sqrt{v_{0}^{2} a^{4} t^{2} - \frac{v_{0}^{2} a^{4} t^{2}}{(1 + a^{2} t^{2})}}$
$a_{N} = a v_{0} \sqrt{\frac{1 + a^{2} t^{2} - a^{2} t^{2}}{(1 + a^{2} t^{2})}}$
$a_{N} = \frac{a v_{0}}{\sqrt{1 + a^{2} t^{2}}}$
$\Rightarrow \frac{a v_{0}}{\sqrt{1 + {(\frac{a y}{v_{0}})}^{2}}}$

Question 3.

A hollow vertical cylinder of radius R and height h has smooth internal surface. A small particle is placed in contact with the inner side of the upper rim at a point P. It is given a horizontal speed v_o tangential to rim. It leaves the lower rim at point Q, vertically below P. The number of revolutions made by the particle will

$\frac{h}{2 π R}$
$\frac{v_{0}}{\sqrt{2 g h}}$
$\frac{2 π R}{h}$
$\frac{v_{0}}{2 π R} (\sqrt{\frac{2 h}{g}})$

Discuss Question

Answer: Option D. ->

\frac{v_{0}}{2 π R} (\sqrt{\frac{2 h}{g}})

:
D

since the body has no initial velocity in the vertical direction.

$a_{z}$ =-g, vertical displacement z=-h.
$∴ z = a t + \frac{1}{2} a t^{2}$
$\Rightarrow - h = 0 + \frac{1}{2} (- g) t^{2}$
$\Rightarrow T = \sqrt{\frac{2 h}{g}}$ time taken to reach the bottom
let,t be the time taken to complete one revolution.
Then $t = \frac{2 π R}{v_{0}}$
$∴$ number of revolution= $\frac{T}{t} = \frac{\sqrt{\frac{2 h}{g}}}{\frac{2 π R}{v_{0}}} = \frac{v_{0}}{2 π R} \sqrt{\frac{2 h}{g}}$

Question 4.

A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of 'P' is such that it sweeps out a length $s = t^{3} + 5$ , where s is in meters and t is in seconds. The radius of the path is 20 m. The acceleration of 'P' when t = 2 s is nearly

13 m/s²
12 m/s²
7.2 m/s²
14 m/s²

Discuss Question

Answer: Option D. -> 14 m/s²
:
D

S = t^{3} + 5

Speed,v=

\frac{d s}{d t} = 3 t^{2}

and rate of change of speed =

\frac{d v}{d t} = 6 t

∴

tangential acceleration at t=2S,

a_{t}

= 6

\times

2=12m/

s^{2}

A t t = 2 s, v = 3 (2)^{2} = 12 m / s^{2}

∴

centripetal acceleration,
=

\frac{v^{2}}{R} = \frac{144}{20} m / s^{2}

∴

net acceleration=

\sqrt{a_{t}^{2} + a_{i}^{2}} \approx 14 m / s^{2}

Question 5.

Diagram shows the direction of the total acceleration and velocity of a particle moving clockwise in a circle of radius 5/√3 m at an instant of time. Tangential acceleration at this instant is 5 m/s². Which of the following statements is not correct?

The centripetal acceleration is $5 \sqrt{3}$ $m / s^{2}$
Particle is speeding up
The net acceleration is 10 $m / s^{2}$
The particle is slowing down

Discuss Question

Answer: Option B. -> Particle is speeding up
:
B and D

since tangential acceleration and v are in same direction, so particle should be speeding up.

$t a n 30^{\circ} = \frac{a_{t}}{a_{c}}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{5}{a_{c}} \Rightarrow$
$a_{c} = 5 \sqrt{3} m / s^{2}$
Net acceleration
$a = \sqrt{a_{c}^{2} + a_{t}^{2}} = \sqrt{(5 \sqrt{3})^{2} + 5^{2}} = 10 m / s^{2}$

Question 6.

A particle moves in a circle of radius 20 cm. Its linear speed is given by v = 2t, where t is second and v in metre/second. Find the radial and tangential acceleration at t = 3 s.

$a_{r} = 1.8 m / s^{2}$
$a_{r} = 180 m / s^{2}$
$a_{t} = 6 m / s^{2}$
$a_{t} = 2 m / s^{2}$

Discuss Question

Answer: Option B. ->

a_{r} = 180 m / s^{2}

:
B and D
The linear speed at t = 3 s is
v = 2t = 6 m/s
The radial acceleration at t = 3 s is

a_{r} = \frac{v^{2}}{r} = 180 m / s^{2}

The tangential acceleration is

a_{t} = \frac{d v}{d t} = 2 m / s^{2}

Question 7.

A car is travelling with speed v on a circular road of radius r. If it is increasing its speed at the rate of 'a' meter/sec², then the resultant acceleration will be

$\sqrt{{\frac{v^{2}}{r^{2}} - a^{2}}}$
$\sqrt{{\frac{v^{4}}{r^{2}} + a^{2}}}$
$\sqrt{{\frac{v^{4}}{r^{2}} - a^{2}}}$
$\sqrt{{\frac{v^{2}}{r^{2}} + a^{2}}}$

Discuss Question

Answer: Option B. ->

\sqrt{{\frac{v^{4}}{r^{2}} + a^{2}}}

:
B

a_{r e s u l t a n t}

\sqrt{a_{r a d i a l}^{2} + a_{t a n g e n t i a l}^{2}} = \sqrt{\frac{v^{4}}{r^{2}} + a^{2}}

Question 8.

Which is these is a possible direction of acceleration for a point on a car that is going on a circular track and is speeding up?

Discuss Question

Answer: Option D. -> d
:
D
The net acceleration is always directed between radially inward and tangential direction.

Question 9.

A body takes the following path and moves with constant speed. If $a_{A}$ and $a_{B}$ are the magnitude of its radial acceleration at A and B.

$a_{A} = a_{B}$
$a_{A} < a_{B}$
$a_{A} > a_{B}$
none of these

Discuss Question

Answer: Option C. ->

a_{A} > a_{B}

:
C

The radius of curvature is the radius of the osculating circle at that point of the curve.

A Body Takes The Following Path And Moves With Constant Spee...

We can clearly see that

R_{A} < R_{B}

∴ \frac{v^{2}}{R_{A}} > \frac{v^{2}}{R_{B}}

(as v is the speed and is constant)

\Rightarrow a_{A} > a_{B}

Question 10.

A particle is projected at angle θ with horizontal with velocity $V_{0}$

Find

(i) tangential and normal acceleration of the particle at t = 0 and at highest point of its trajectory.

(ii) radius of curvature at t = 0 and highest point.

at $t = 0, a_{t} = - g s i n θ$
At highest point $a_{n} = - g$
at $t = 0, R = \frac{V^{2}}{s i n θ}$
At highest point $R = \frac{V^{2} c o s^{2} θ}{g}$

Discuss Question

Answer: Option C. -> at

t = 0, R = \frac{V^{2}}{s i n θ}

:
A, B, C, and D

(i) The direction of tangential acceleration is in the line of velocity and the normal acceleration is in the direction perpendicular to velocity direction. The tangential and normal directions at O & P are shown in figure.
The net acceleration of the particle during motion is acceleration due to gravity i.e., 'g' is acting vertically downward.

At O:(at t=0)
Tangential acceleration $a_{t}$ =- $g s i n θ$
Normal acceleration $a_{n}$ = $g c o s θ$
At P (at highest point)
Tangential acceleration
$a_{t} = 0 a_{n} = g$
(ii) Let radius of curvature at O be $R_{0}$ . The normal acceleration at O is g cos $θ$ .
$a_{n} = \frac{v^{2}}{R} \Rightarrow R_{0} = \frac{v_{0}^{2}}{a_{n}} = \frac{v_{0}^{2}}{g c o s θ}$

Radius of curvature at P
Normal acceleration at P is 'g'
$R_{p} = \frac{v^{2}}{a_{n}} = \frac{(v_{0} c o s θ)^{2}}{g} = \frac{v_{0}^{2} c o s^{2} θ}{g}$