## KINEMATICS MCQs

### Circular Kinematics

Total Questions : 41 | Page 3 of 5 pages
Question 21. A balloon starts rising from the surface of the Earth. The ascension rate is constant and equal to vo. Due to the wind the balloon gathers the horizontal velocity component vx=ay, where a is a constant and y is the height of ascent. Find the total, tangential, and normal accelerations of the balloon.

1.    ar=av0;at=av0√1+(ayv0)2;aN=av0
2.    ar=av0;at=av0;aN=0
3.    ar=av0;at=a2y√1+a2y2v20;aN=av0 ⎷(1+a2y2v20)
4.    ar=a2y√1+a2y2v20;at=av0;aN=av0
Answer: Option C. -> ar=av0;at=a2y√1+a2y2v20;aN=av0 ⎷(1+a2y2v20)
:
C
The path of the balloon will look something like this

After t sec ballon would have gone a height of v0tthen at that very instance the balloon's vx will be
vx=av0tax=av0
Vy=v0
ay=0
ar=totalacceleration=av0.......(1)
vtangentical is actually the resultant velocity
vt=v20+(av0t)2
vt=v01+a2t2
at=dvtdt=v0a2t1+a2t2a2y1+(ayv0)2
Now we know a2t+a2N=a2r
v20a4t2(1+a2t2)+a2N=a2v20
aN=v20a4t2v20a4t2(1+a2t2)
aN=av01+a2t2a2t2(1+a2t2)
aN=av01+a2t2
av01+(ayv0)2
Question 22. 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 vo tangential to rim. It leaves the lower rim at point Q, vertically below P. The number of revolutions made by the particle will
1.    h2πR
2.    v0√2gh
3.    2πRh
4.    v02πR(√2hg)
Answer: Option D. -> v02πR(√2hg)
:
D
since the body has no initial velocity in the vertical direction.

az=-g, vertical displacement z=-h.
z=at+12at2
h=0+12(g)t2
T=2hg time taken to reach the bottom
let,t be the time taken to complete one revolution.
Then t=2πRv0
number of revolution=Tt=2hg2πRv0=v02πR2hg
Question 23. Find the maximum velocity for skidding for a car moved on a circular track of radius 100 m. The coefficient of friction between the road and tyre is 0.2
1.    0.14 m/s
2.    140 m/s
3.    1.4 km/s
4.    14 m/s
Answer: Option D. -> 14 m/s
:
D
Vmax=μrg=0.2×100×9.8=14m/s
Question 24. 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=t3+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
1.    13 m/s2
2.    12 m/s2
3.    7.2 m/s2
4.    14 m/s2
Answer: Option D. -> 14 m/s2
:
D
S=t3+5
Speed,v=dsdt=3t2 and rate of change of speed =dvdt=6t
tangential acceleration at t=2S,at = 6 × 2=12m/s2
Att=2s,v=3(2)2=12m/s2
centripetal acceleration,
=v2R=14420m/s2
net acceleration=a2t+a2i14m/s2
Question 25. 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?
1.    a
2.    b
3.    c
4.    d
Answer: Option D. -> d
:
D
The net acceleration is always directed between radially inward and tangential direction.
Question 26. A particle of mass M moves with constant speed along a circular path of radius r under the action of a force F.  Its speed is
1.    √rFm
2.    √Fr
3.    √Fmr
4.    √Fmr
Answer: Option A. -> √rFm
:
A
F=mv2rv=rFm
Question 27. A body takes the following path and moves with constant speed. If aA and aB are the magnitude of its radial acceleration at A and B.
1.    aA=aB
2.    aA
3.    aA>aB
4.    none of these
Answer: Option C. -> aA>aB
:
C
The radius of curvature is the radius of the osculating circle at that point of the curve.

We can clearly see that RA<RB
v2RA>v2RB (as v is the speed and is constant)
aA>aB
Question 28. The speed of the truck is 40 m s-1, after 10 seconds its speed decreases to 20 m s-1, its acceleration is
1.    -1m/s2
2.    -2m/s2
3.    -4m/s2
4.    -5m/s2
Answer: Option B. -> -2m/s2
Question 29. Which of the following is NOT a unit for acceleration?
1.    m/s2
2.    cm/s2
3.    km/hr
4.    km/hr2
Answer: Option C. -> km/hr
Question 30. Calculate the speed of a sprinter who ran 200 meters in 20 seconds
1.    10m/s
2.    10km/s
3.    10m/hr
4.    10km/hr
Answer: Option A. -> 10m/s