12th Grade > Physics
KINEMATICS MCQs
Circular Kinematics
Total Questions : 41
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Answer: Option C. -> constant kinetic energy
:
C
![Click to Enlarge Image A Body Moving In A Circular Path With A Constant Speed Has A...](https://lakshyaeducation.in/quizpics/quiz/59365_0cae54bcf9241e1291ab4d9231d8617b910de31820160701-5603-17j35qu.png)
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.
∴ Answer is kinetic energy.
:
C
![Click to Enlarge Image A Body Moving In A Circular Path With A Constant Speed Has A...](https://lakshyaeducation.in/quizpics/quiz/59365_0cae54bcf9241e1291ab4d9231d8617b910de31820160701-5603-17j35qu.png)
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.
∴ Answer is kinetic energy.
Answer: Option A. -> 40π2
:
A
Att1=2s,→a=(5√3ms2)^i+5^j
Att2=5s,→a=(5ms2)^i−(5√3ms2)^i+^j
Given the body is moving in anti-clockwise direction
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_a8718af38593777fe50149487e66032f3c6814b020160609-1956-jq2vma.png)
Since,the speed is constant, net acceleration will always be directed toward the centre and since it is
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_b229d2a20569860e367e32eae404ccb91fc7d76320160609-1956-1nhmv7f.png)
This will be the only path possible using this, we can calculate ω as we can find the change in angle
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_91e45806fcc6f8a305461535ed7f5800fe95062b20160609-1956-1vkz8vw.png)
tanθ1=15√3=13
tanθ2=3√35=√3
∴θ1=30∘
⇒θ2=60∘
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_359d0651e26092a2a8bf5b2b28b67b52444bff4f20160609-1956-13l6813.png)
∴ We can see that angle between the vectors is 90∘ which means particle would have covered an angle of 360−90=270∘=3π/2radin3s
∴ω=ΔθΔt=3π23=π2rad/s
a=ω2r
r=aω2=√(5√3)2+52(π2)2=√100π24
=4×10π2=40π2
:
A
Att1=2s,→a=(5√3ms2)^i+5^j
Att2=5s,→a=(5ms2)^i−(5√3ms2)^i+^j
Given the body is moving in anti-clockwise direction
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_a8718af38593777fe50149487e66032f3c6814b020160609-1956-jq2vma.png)
Since,the speed is constant, net acceleration will always be directed toward the centre and since it is
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_b229d2a20569860e367e32eae404ccb91fc7d76320160609-1956-1nhmv7f.png)
This will be the only path possible using this, we can calculate ω as we can find the change in angle
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_91e45806fcc6f8a305461535ed7f5800fe95062b20160609-1956-1vkz8vw.png)
tanθ1=15√3=13
tanθ2=3√35=√3
∴θ1=30∘
⇒θ2=60∘
![Click to Enlarge Image At T1 = 2.00 S, The Acceleration Of A Particle In Counter ...](https://lakshyaeducation.in/quizpics/quiz/59388_359d0651e26092a2a8bf5b2b28b67b52444bff4f20160609-1956-13l6813.png)
∴ We can see that angle between the vectors is 90∘ which means particle would have covered an angle of 360−90=270∘=3π/2radin3s
∴ω=ΔθΔt=3π23=π2rad/s
a=ω2r
r=aω2=√(5√3)2+52(π2)2=√100π24
=4×10π2=40π2
Answer: Option D. -> 1
:
D
ω1=△θ△t
If time taken to complete a revolution is t.
Then ω1=2πt,ω2=2πt
⇒t=2πω1=2πω2
⇒ we get ω1=ω2orω1ω2=1
:
D
ω1=△θ△t
If time taken to complete a revolution is t.
Then ω1=2πt,ω2=2πt
⇒t=2πω1=2πω2
⇒ we get ω1=ω2orω1ω2=1
Answer: Option B. -> Q
:
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.
:
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.
Answer: Option B. -> R1
:
B
Reaction on inner wheel R1=12M[g−v2hra]
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.
:
B
Reaction on inner wheel R1=12M[g−v2hra]
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.
Answer: Option C. -> Tangentially outward
:
C
Stone flies in the direction of instantaneous velocity due to inertia
:
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
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
Answer: Option C. -> 3 ^i + 6 ^j
:
C
![Click to Enlarge Image A Purse At Radius 2.00 M And A Wallet At Radius 3.00 M Tra...](https://lakshyaeducation.in/quizpics/quiz/content_10.png)
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=3√5
So the above answer matches with the magnitude of third option in the given answers.
:
C
![Click to Enlarge Image A Purse At Radius 2.00 M And A Wallet At Radius 3.00 M Tra...](https://lakshyaeducation.in/quizpics/quiz/content_10.png)
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=3√5
So the above answer matches with the magnitude of third option in the given answers.
Answer: Option B. -> √{v4r2+a2}
:
B
aresultant=√a2radial+a2tangential=√v4r2+a2
:
B
aresultant=√a2radial+a2tangential=√v4r2+a2
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
:
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?
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?
Answer: Option A. -> (15 m, 6 m)
:
A
We know the body is undergoing uniform circular motion,
![Click to Enlarge Image A Particle Moves Along A Circular Path Over A Horizontal X...](https://lakshyaeducation.in/quizpics/quiz/59381_4c762788471d9fe61d75a295db2b84670b5712d820160610-24159-18hmhn0.png)
∴ only possible path is
![Click to Enlarge Image A Particle Moves Along A Circular Path Over A Horizontal X...](https://lakshyaeducation.in/quizpics/quiz/59381_f2ccf5668a0386b755f38adfe40e5411213a173a20160610-24159-yv4qyt.png)
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=3π10S−1
usingv=ωr
r=vω
|→v|=3πm/sgiven
∴r=3π3π10=30π3π=10m
∴ifr=10m, coordinates of the centre (15 m,6m)
:
A
We know the body is undergoing uniform circular motion,
![Click to Enlarge Image A Particle Moves Along A Circular Path Over A Horizontal X...](https://lakshyaeducation.in/quizpics/quiz/59381_4c762788471d9fe61d75a295db2b84670b5712d820160610-24159-18hmhn0.png)
∴ only possible path is
![Click to Enlarge Image A Particle Moves Along A Circular Path Over A Horizontal X...](https://lakshyaeducation.in/quizpics/quiz/59381_f2ccf5668a0386b755f38adfe40e5411213a173a20160610-24159-yv4qyt.png)
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=3π10S−1
usingv=ωr
r=vω
|→v|=3πm/sgiven
∴r=3π3π10=30π3π=10m
∴ifr=10m, coordinates of the centre (15 m,6m)