1

Chapter 01 : Circular Motion

1.

N = 2mv

mgsinR

For equilibrium,

mg cos = N = 2mv

mgsinR

….(i)

From energy conservation,

1

2mv2 = mg R (sin )

2mv

R= 2 mg sin ….(ii)

mg cos = (2 mg sin + mg sin )

….[From (i) and (ii)]

= cos

3sin

tan = 1

3

= 45 …. 1

3

2. r = b

sin

v = v sin

Now, = v/r

= vsin

bsin

= 2vsin

b

3. TT is the tangent to the curve at point P. mg sin = (m 2x) cos ….[along TT]

tan = 2x

g

dy

dx=

2x

g

But,

dy

dx= 3 4d

a xdx

= 4 a3x3

4 a3x3 = 2x

g

= 2x 3a g

4. AB A Bv v v

Now,

ABv

= 2 2 2v v 2v cos(180 )

[smaller angle between Av

and Bv

= 180 ]

= 22v (1 cos )

= 2 22v 2sin ( / 2)

= 2 v sin (/2)

= 2 R sin (/2) 5. Since this is not a case of a normal string, the

velocity at the topmost point can be zero. (T.E.)initial = (T.E.)final

mgh + 1

2mv2 = mg (2R)

v = 2g(2R h)

R

mg

NN

R

2mv

R

Y

X

90 vP

(0, b)

r

Y

X

m 2r

T

P(x, y)mg

N

T

B

A

Bv

Av

Circular Motion01 

2

Std. XII : Triumph Physics 

Note: In case of a string, v at the topmost point

should be equal to Rg to complete the

vertical circle as T = 0 and ball will fall vertically down if v = 0.

6. P.E. = mg R (1 – cos ) and

K.E. = 1

2mv2

(Work done)pseudo force = mgR sin

mg R (1 – cos ) + mg R sin = 1

2mv2

mg R (1 – cos + sin ) = 1

2mv2

v = 2gR (1 cos sin ) 7.

tan 60 = r

t

a

a

ar = ta 3

2v

r = ta 3 ….(i)

v = area under graph.

v = ta t

2 ….(ii)

2 2ta t

4(1) = ta 3 ….[From (i) and (ii)]

2

ta .t3

4 ….(iii)

Also, tan (60) = ta

t

3 = ta

t or ta t 3 ….(iv)

3t 3

4 = 3 ….[From (iii) and (iv)]

t3 = 4 t = 2/32 s

8. ar =

2

2

2

vvsin

R R sin

…[ vt = v/sin ]

Also, R(1 cos )

tv

cos = vt

1R

ar =2tv

R

2

2

v

vtR 1 1

R

= 2

2 2

2

v

2vt v tR

R R

= 2

Rv

2Rt vt

9. mgh = 21mv

2

v = 2gh

cos = h

l

T = 2mv

mgcosr

T = 2mgh h 3mg

mg h l l l

which implies a straight line graph. 10.

favg = 0

(Ncos )

Here, integration is not possible. So, we use the fact that we need to calculate

favg

favg = p

t

Favg = (2mv)

r

v

= 22mv

r

11. N cos = 20mv

r and N sin = mg

tan = 20

g

v

r

r = 2

0vtan

g

60at

ar

rN

mg

20mv

r

v v

sin

at

t

60

mg

2mv

r

lh

T

N cos

FBD of tubeFBD of the ball

mg

2mv

R

N

3

Chapter 01 : Circular Motion

12. Angle moved = in time t

t = v

l ….(v = velocity of bullet)

Also, = t

=v

l v =

l

13. 1

d dk kt c

dt dt

= 1(kt c )dt

= 2

1 2

ktc t c

2

= quadratic equation which has a graph of parabola

14. Friction will act in upward direction. Since velocity is a constant,

N = 2mv

mgsinR

f = 2mv

mgsin mgcosR

[at = 0]

As increases, cos decreases friction decreases.

Again, at = 0

Friction = 2mv

mgsinR

= mg cos

As decreases, cos increases friction increases.

15. The area under the t graph gives change in

angular velocity.

Area = 2(2) 4

2 2

= 2

2 1 = 2 2 = 2 + 2 = 4 rad/s

16. Velocity is a vector which changes but speed remains same for uniform circular motion.

In case A, radius of curvature remains same

throughout hence a = 2v

r remains constant.

However, in case of B, the radius of curvature

keeps increasing hence a = 2v

r keeps

decreasing. Hence option (C) is the only correct option.

17. The direction of rotation is determined by the

sign of angular velocity. In turn, the sign of angular velocity is determined by the sign of slope on angular displacement vs time plot. The sign of slope is negative for line OA, positive for line AC and zero for line CD.

The positive angular velocity indicates anti-clockwise rotation and negative angular velocity indicates clockwise rotation. The disk is stationary when angular velocity is zero.

18. m2r cos = mg sin

2 = g tan

r

tan = 2v

rg

21000 m

723600 sh

(400m)(10m/s)

l

h 1

1m 10

h = 10 cm 19. At the highest point,

= g

R= 2n

n = 2

1 g g

2 R 4 R

r.p.m. = 60n = 602 2

g 900g

4 n R

20. = d

d

So is negative, if

> 0, d

0d

or < 0,

d0

d

BA l

f

2mv

R

mg

N

f

2mv

R

mg

N

mg cos mg sin

N

m2r

h

mg

l

4

Std. XII : Triumph Physics 

21. For option (A),

Net force = Mv2/r = Mass acceleration For option (B),

ta

and

are perpendicular hence cross product is not 0.

For option (C), Angular velocity and angular accleration have

the same direction or opposite direction according to the type of motion.

For option (D), The correct statement is: The resultant force acts always towards the

centre. 22. Weight = Number of balls centripetal force

(400) (10) = 8 m 2r

= 8 (5) 2 (1)

2 = 4000

40

= 100

= 10 rad/s 23. Take a small mass element dm This element experiences a centripetal force

along radial direction,

Fd = 2v

(dm)R

The components T cos d

2

cancel each other

2T sind

2

= (dm) 2v

R

T d = 2M v

Rd2 R R

sin

as 0

T = 2Mv

2 R

24.

Energy conservation, mgl(1 cos) = 1

2mv2

v = 2g (1 cos ) l

2v

l= g sin

2g (1 cos ) = g sin

2(1 cos) = sin

2 = 2

2sin cossin 2 2

1 cos2sin

2

cot2

= 2

i.e. = 53 25. = a(t2) i + b(et) j

= d

dt

= 2a(t) i + (b)(et) j

at t = 1 s and = bjˆaie

= bjˆ2aie

= 2a2 2

2

b

e =

2 22 2

2 2

b ba 4a

e e cos

= cos1

22

2

2 22 2

2 2

b2a

e

b ba 4a

e e

30 ….[ a = b = 1]

R

C

M

TT

Tcos d

2

Tcos d

2

(dm)2v

R

Tsin d

2

Tsin d

2

C

d

400 kg

m2rr

m

2v

l

l

gv m

l

gg sin