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 ___
1.    constant velocity
2.    constant momentum
3.    constant kinetic energy
4.    constant acceleration
Answer: Option C. -> constant kinetic energy
:
C

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 12m|v|2 where|v| is the speed.
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?
1.    40π2
2.    360π2
3.    30π2
4.    60π2
:
A
Att1=2s,a=(53ms2)^i+5^j
Att2=5s,a=(5ms2)^i(53ms2)^i+^j
Given the body is moving in anti-clockwise direction

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

tanθ1=153=13
tanθ2=335=3
θ1=30
θ2=60

We can see that angle between the vectors is 90 which means particle would have covered an angle of 36090=270=3π/2radin3s
a=ω2r
r=aω2=(53)2+52(π2)2=100π24
=4×10π2=40π2
Question 3. Two cars having masses m1 and m2 move in circles of radii r1 and r2 respectively. If they complete the circle in equal time, the ratio of their angular speeds ω1ω2 is _______
1.    m1m2
2.    r1r2
3.    m1r1m2r2
4.    1
:
D
ω1=θt
If time taken to complete a revolution is t.
Then ω1=2πt,ω2=2πt
t=2πω1=2πω2
we get ω1=ω2orω1ω2=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?
1.    P
2.    Q
3.    both have equal
4.    can't say until value of ω is given
:
B
If a particle goes in a circle with constant ω then it has a centripetal acceleration given by ω2R since ω is constant for both particles. So the acceleration depends on R
RQ>RP
ω2RQ>ω2RP
So particle q will have higher centripetal acceleration.
Question 5. A car is moving on a circular path and takes a turn. If R1 and R2  be the reactions on the inner and outer wheels respectively, then
1.    R1 = R2
2.    R1
3.    R1 > R2
4.    R1 ≥ R2
:
B
Reaction on inner wheel R1=12M[gv2hra]
Reaction on outer wheel R2=12M[g+v2hra]
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
3.    Tangentially outward
4.    With an acceleration mv2l
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/ s2)^i    + (4.00 m/ s2)^j    . At that instant and in unit-vector notation, what is the acceleration of the wallet?                                                                   IIT JEE- 2001
1.    2 ^i  + 4 ^j
2.    4 ^i  + 2 ^j
3.    3 ^i  + 6 ^j
4.    3 √5 ^i  + 3 √5 ^j
Answer: Option C. -> 3 ^i  + 6 ^j
:
C

dθdt is constant.
In other words in uniform circular motion the angular velocity remains constant body doesn't have any tangential acceleration but normal acceleartion.
aN=v2Rorω2R
ForpurseaN=(2)2+(4)2=20;R=2
20=ω22
ω2=5
ForwalletaN=ω2R
Hence ωis same
But~R=3
aN=5×3
aN=35
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
1.    √{v2r2−a2}
2.    √{v4r2+a2}
3.    √{v4r2−a2}
4.    √{v2r2+a2}
:
B
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?
1.    10 π m / minute
2.    20 π m / minute
3.    30 π m / minute
4.    5 π m / minute
Answer: Option B. -> 20 π m / minute
:
B
The linear speed is,
v = r ω ;ω = θt
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?
1.    (15 m, 6 m)
2.    (-5 m, 6)
3.    (15 m, 0)
4.    (-5 m, 0)
Answer: Option A. -> (15 m, 6 m)
:
A
We know the body is undergoing uniform circular motion,

only possible path is

As acceleration should always be directed toward the centre
To calculate the position of the centre, we need the radius, to calculate radius,v=ωr we can use the relation,as we can calculate ω
Δθ=3π2,Δt=5s
=3π2×5=3π10S1
usingv=ωr
r=vω
|v|=3πm/sgiven
r=3π3π10=30π3π=10m
ifr=10m, coordinates of the centre (15 m,6m)