Kinematics(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

KINEMATICS MCQs

Circular Kinematics

Total Questions : 41 | Page 1 of 5 pages

Question 1. A body moving in a circular path with a constant speed has a ___

constant velocity
constant momentum
constant kinetic energy
constant acceleration

Discuss Question

Answer: Option C. -> constant kinetic energy
:
C

A Body Moving In A Circular Path With A Constant Speed Has A...

We can see that direction of velocity and acceleration changes continuously

∴

Both are not constant.
Momentum is m.

⃗ v

which depends on velocity and changes with velocity.
The only quantity that is constant is speed and kinetic energy is

\frac{1}{2} m | ⃗ v |^{2}

where

| ⃗ v |

is the speed.

∴

Answer is kinetic energy.

Question 2.
At t₁ = 2.00 s, the acceleration of a particle in counter clockwise circular motion is

It moves at constant speed. At time t₂ = 5.00 s, the particle's acceleration is

What is the radius of the path taken by the particle if t₂ - t₁ is less than one period?

40π2
360π2
30π2
60π2

Discuss Question

Answer: Option A. -> 40π2
:
A

A t t_{1} = 2 s, \to a = (5 \sqrt{3} \frac{m}{s^{2}})^i + 5^j

A t t_{2} = 5 s, \to a = (5 \frac{m}{s^{2}})^i - (5 \sqrt{3} \frac{m}{s^{2}})^i +^j

Given the body is moving in anti-clockwise direction

At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...

Since,the speed is constant, net acceleration will always be directed toward the centre and since it is

This will be the only path possible using this, we can calculate ω as we can find the change in angle

t a n θ_{1} = \frac{1}{5 \sqrt{3}} = \frac{1}{3}

t a n θ_{2} = \frac{3 \sqrt{3}}{5} = \sqrt{3}

∴ θ_{1} = 30^{\circ}

\Rightarrow θ_{2} = 60^{\circ}

∴

We can see that angle between the vectors is

90^{\circ}

which means particle would have covered an angle of

360 - 90 = 270^{\circ} =^{3 π / 2} r a d i n 3 s

∴ ω = \frac{Δ θ}{Δ t} = \frac{\frac{3 π}{2}}{3} = \frac{π}{2} r a d / s

a = ω^{2} r

r = \frac{a}{ω^{2}} = \frac{\sqrt{(5 \sqrt{3})^{2}} + 5^{2}}{{(\frac{π}{2})}^{2}} = \frac{\sqrt{100}}{\frac{π^{2}}{4}}

\frac{4 \times 10}{π^{2}} = \frac{40}{π^{2}}

Question 3. Two cars having masses

m_{1} a n d m_{2}

move in circles of radii

r_{1} a n d r_{2}

respectively. If they complete the circle in equal time, the ratio of their angular speeds

\frac{ω_{1}}{ω_{2}}

is _______

m1m2
r1r2
m1r1m2r2
1

Discuss Question

Answer: Option D. -> 1
:
D

ω_{1} = \frac{△ θ}{△ t}

If time taken to complete a revolution is t.
Then

ω_{1} = \frac{2 π}{t}, ω_{2} = \frac{2 π}{t}

\Rightarrow t = \frac{2 π}{ω_{1}} = \frac{2 π}{ω_{2}}

\Rightarrow

we get

ω_{1} = ω_{2} o r \frac{ω_{1}}{ω_{2}} = 1

Question 4.
A circular disc is rotating with constant angular velocity ω about an axis that passes through the centre. A particle 'P' is kept at a distance of 2m from the centre and another particle 'Q' at a distance of 3m from the centre.
Which of these will have higher centripetal acceleration?

P
Q
both have equal
can't say until value of ω is given

Discuss Question

Answer: Option B. -> Q
:
B
If a particle goes in a circle with constant ω then it has a centripetal acceleration given by

ω^{2} R

since ω is constant for both particles. So the acceleration depends on R

∵ R_{Q} > R_{P}

\Rightarrow ω^{2} R_{Q} > ω^{2} R_{P}

So particle q will have higher centripetal acceleration.

Question 5. A car is moving on a circular path and takes a turn. If

R_{1} a n d R_{2}

be the reactions on the inner and outer wheels respectively, then

R1 = R2
R1
R1 > R2
R1 ≥ R2

Discuss Question

Answer: Option B. -> R1
:
B
Reaction on inner wheel

R_{1} = \frac{1}{2} M [g - \frac{v^{2} h}{r a}]

Reaction on outer wheel

R_{2} = \frac{1}{2} M [g + \frac{v^{2} h}{r a}]

where, r = radius of circular path, 2a = distance between two wheels and h = height of centre of gravity of car.

Question 6. A stone of mass m is tied to a string of length l and rotated in a circle with a constant speed v. If the string is released, the stone flies

Radially outward
Radially inward
Tangentially outward
With an acceleration mv2l

Discuss Question

Answer: Option C. -> Tangentially outward
:
C
Stone flies in the direction of instantaneous velocity due to inertia

Question 7.
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 √5 ^i + 3 √5 ^j

Discuss Question

Answer: Option C. -> 3 ^i + 6 ^j
:
C

A Purse At Radius 2.00 M And A Wallet At Radius 3.00 M Tra...

\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 8. 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

√{v2r2−a2}
√{v4r2+a2}
√{v4r2−a2}
√{v2r2+a2}

Discuss Question

Answer: Option B. -> √{v4r2+a2}
:
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 9. A car moves on a circular path of radius 10 metre. It complete 5 revolution in 5 minute. What is its average speed?

10 π m / minute
20 π m / minute
30 π m / minute
5 π m / minute

Discuss Question

Answer: Option B. -> 20 π m / minute
:
B
The linear speed is,
v = r

ω

;

ω

\frac{θ}{t}

Substituting, we get,
v=20

π

m / minute

Question 10.
A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t₁ = 5.00 s, it is at point (5.00 m, 6.00 m) with velocity (3π m/s)

^j

and acceleration in the positive x direction. At time t₂ = 10.0 s, it has velocity (-3π m/s)

^i

and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path if t₂ - t₁ is less than one period?