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

12th Grade > Physics

KINEMATICS MCQs

Circular Kinematics

Total Questions : 41 | Page 2 of 5 pages

Question 11.
Assume that the earth goes round the sun in a circular orbit with a constant speed of 30 km/s. Then,

The average velocity of the earth from 1st Jan, 90 to 30th June, 90 is zero
The average acceleration during the above period is 60 km/s2
The average speed from 1st Jan, 90 to 31st Dec, 90 is zero
The instantaneous acceleration of the earth points towards the sun.

Discuss Question

Answer: Option D. -> The instantaneous acceleration of the earth points towards the sun.
:
D

Assume That The Earth Goes Round The Sun In A Circular Orb...

(a) we can see there is a displacement from A to B .

∴

average velocity cannot be zero
(b)

a_{a v g} = \frac{{\to v}_{2} - {\to v}_{1}}{t} = \frac{30^i - (- 30^i)}{6 m o n t h s} = \frac{60}{6 \times 30 \times 3600} k m / s^{2}

which is already not 60 km/

s^{2}

(c) average speed is not zero as total distance covered is not zero
(d) The instantaneous acceleration will point to the sun as the above described motion is uniform circular motion with sun at the centre of the circular path.

Question 12. Consider two masses

m_{1}

and

m_{2}

are moving in circles of radii

r_{1}

and

r_{2}

respectively. Their speeds are such that they complete circular motion in the same time t. The ratio of their untripetal acceleration is,

m1 : m2
r1 : r2
1 : 1
m1 r1 : m2r2

Discuss Question

Answer: Option B. -> r1 : r2
:
B
Centripetal acceleration,
a =

w^{2}

r =

(\frac{2 π}{T})^{2}

. r

∴

\frac{a_{1}}{a_{2}}

\frac{r_{1}}{r_{2}}

Question 13.
Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.

0.0336 m/s2
4π2m/s2
336 m/s2
none of these

Discuss Question

Answer: Option A. -> 0.0336 m/s2
:
A
Earth completes one rotation in 24 hrs
i.e. angle swept =2

π

in time=24 hrs

\Rightarrow a n g u l a r v e l o c i t y = ω = \frac{2 π}{24 \times 60 \times 60}

Centripetal acceleration =

ω^{2} R

\Rightarrow {(\frac{2 π}{24 \times 60 \times 60})}^{2} \times 6400 \times 1000

(∵ R = \frac{D i a m e t e r}{2})

\Rightarrow 0.0336 m / s^{2}

Question 14.
A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3 m/s and a centripetal acceleration

\to a

of magnitude 2 m/s². Position vector

\to r

locates him relative to the rotation axis.
What is the magnitude of

\to r

? What is the direction of

\to r

when

\to a

is directed due east?

1.5 m, east
4.5 m, east
1.5 m, west
4.5 m, west

Discuss Question

Answer: Option D. -> 4.5 m, west
:
D

v = 3 m / s, a = 2 m / s^{2}

a = \frac{v^{2}}{r} \Rightarrow r = \frac{v^{2}}{a} = \frac{3^{2}}{2} = \frac{9}{2} = 4.5 m

∴ | \to r | = 4.5 m

When

\to a

is due east

A Carnival Merry-go-round Rotates About A Vertical Axis At...

\to r

will be due west as r points to the particle's position W.r.t. centre
As

\to a

always points toward the centre of motion
Please watch the video if you are still doubtful about why a:

\frac{v^{2}}{r}

Question 15. A circular disc is rotating with constant

ω = 2

rad/sec, about an axis passing through its centre. An object is kept at a distance of

^{'} r_{1} = 2 c m^{'}

from the centre and another object is kept at a distance of

^{'} r_{2} = 1 c m^{'}

. Find the ratio of their linear speeds.
___

Discuss Question

v = r ω

v_{1} = r_{1} ω = 2 \times 2 = \frac{4 c m}{s e c}

v_{2} = r_{2} ω = 1 \times 2 = \frac{2 c m}{s e c}

\frac{v_{1}}{v_{2}} = \frac{4}{2} = 2

Question 16. A particle moves in a circle of radius 20cm with a linear speed of 10 m/s. Find the angular velocity, in rad/sec is
___

Discuss Question

:
The angular velocity is

ω = \frac{v}{r} = \frac{10 \frac{m}{s}}{20 c m} = 50 \frac{r a d}{s}

Question 17. A car is going with constant speed on a circular path. If it starts from point A at t = 0 and reaches diametrically opposite point B after 2 sec then find its angular velocity.

π
π2
3π2
2π

Discuss Question

Answer: Option B. -> π2
:
B

A Car Is Going With Constant Speed On A Circular Path. If It...

In going from A to B the car is displaced by an angle of

180^{\circ}

.
So, angular displacement =

π

radians
Time = 2 sec
Angular velocity =

\frac{a n g u l a r d i s p l a c e m e n t}{t i m e}

ω = \frac{π}{2} \frac{r a d}{s e c}

Question 18. For a particle in uniform circular motion the acceleration

\to a

at a point

P (R, θ)

on the circle of radius R is (here

θ

is measured from the x-axis)

For A Particle In Uniform Circular Motion The Acceleration �...

−v2Rcosθ^i+v2Rsinθ^j
−v2Rsinθ^i+v2Rcosθ^j
−v2Rcosθ^i+v2Rsinθ^j
v2R+v2R^j

Discuss Question

Answer: Option C. -> −v2Rcosθ^i+v2Rsinθ^j
:
C

\to r = R c o s θ^j

∴ \frac{d \to r}{d t} = - R s i n θ \frac{d θ}{d t}^i + R c o s θ \frac{d θ}{d t}^j

\Rightarrow \to v = - R ω s i n θ^i + R ω c o s θ^j

(a s ω = \frac{d θ}{d t} a n d \frac{d \to r}{d t} = \to v)

\Rightarrow \frac{d \to v}{d t} = \to a = R ω c o s θ \frac{d θ}{d t}^i + R ω (- s i n θ) \frac{d θ}{d t}^j

\Rightarrow \frac{d s}{d t} \equiv \frac{d θ r}{d t} = r \frac{d θ}{d t} = r ω

∴ ω o r \frac{d θ}{d t} i s c o n s t a n t

∴ \to v = - R ω^{2} c o s θ^i - R ω^{2} s i n θ^j

Substituting

ω = \frac{v}{R}

\to a = \frac{- v^{2}}{R} c o s θ^i - \frac{- v^{2}}{R} s i n θ^j

Question 19. A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point at coordinates (4.00 m, 4.00 m) with a velocity of -5.00

^i

m/s and an acceleration of + 12.5

^j

m/s². What are the x and y coordinates of the center of the circular path?

(4, 6)
(4, 2)
(2, 4)
(6, 4)

Discuss Question

Answer: Option A. -> (4, 6)
:
A

A Particle Moves Horizontally In Uniform Circular Motion, Ov...

since

a_{c}

points in y direction that means centre should be right above it
We know

a_{c}

\frac{v^{2}}{R}

12.5 = \frac{(2)^{2}}{R}

\Rightarrow R = 2

\Rightarrow

The centre is 2 units above the point(4,4)

\Rightarrow

Coordinates of centre =(4,6)

Question 20. A particle moves with constant speed v along a circular path of radius r and completes the circle in time T. The acceleration of the particle is