If speed is constant angle between v → and a → will always be 90°

av B A

tAB

= BD→ t_AB

i.e., | | | |

a BD

→ =

→

=

av t

v

AB tAB

2

Now, | | | |

v AB

→ =

→

=

av t R

AB tAB

2

∴ | |

| |

a v

→

→^av = =

av

v R

2

2 ω (angular velocity) Thus, assertion is correct.

In circular motion, when speed is constant, the angular velocity will obviously be constant; but this reason does not lead to the result as explained.

Option (b) is correct.

3. A frame moving in a circle with constant speed can never be an inertial frame as the frame is not moving with constant velocity (due to change in direction).

Reason that the frame is having constant acceleration is false.

Option (c) is correct.

4. If speed is constant angle between v^→ and a^→ will always be 90°

∴ ^{→ →}v a⋅ = 0

B

D

A vB

vA

→

→ v→A

2.5 ms–1

B

C A–1

1 ms 2 ms^–1

a = a_c

ω θ

→

→ →

If speed is increasing angle θ between v^→ and a^→ will be less than 90°.

∴ ^{→ →}v a⋅ will be positive.

If speed is decreasing angle between v^→ and a→

will be greater than 90°.

∴ v a^{→ →}⋅ will be negative.

Assertion is correct.

Rea son ω^{→ →}⋅v=0 as both are perpendicular to each other.

Reason is also true but not the correct explanation of the assertion.

Option (b) is correct.

5. ^→v= 2^{^}i ms⁻¹

a^→= −^{^}i+2 ms^{^}j ⁻²

∴ a^→c =2^{^}jms⁻² a^→_c =| | 2 msa^→_c = ⁻² v =| |^→v =2 ms⁻¹ a v

c = R

2

⇒ R v a_c

=

2

=( )2 =

2 2

2

m a^→_T = −^{^}i ms⁻²

Speed is decreasing (as a^→_T is −ive) at a rate of 1 ms⁻¹ per second i.e., 1 ms⁻². Both Assertion and the Reason are correct but reason has nothing to do with the assertion.

Option (b) is correct.

6. a^→=a^→_T +^→a_c (Reason)

∴ | |^→a = v + r g

2 2

⇒ | |^→a > g (Assertion)

We see that both Assertion and the Reason are correct and the reason is the correct explanation of assertion.

Option (a) is correct.

7. At points A and C : The bob is momentarily at rest.

i.e., v = 0 (Reason)

∴ |a_c| v R

→ = =

2

0

but net acceleration is not zero (see figure) i.e., Assertion is false.

∴ Option (d) is correct.

Circular Motion |

151

ac

θ

v a_T

a

→ →

→ →

R

2j v = 2z a_net

i

x

–

→ a_c θ v aT

a

→ →

→ →

a = g_T a_c

a

A

→ →

→

B

g g

A C

→ →

8. v (speed) =4t−12

For t < 3 (time unit) speed is negative, which can’t describe a motion. Thus, assertion is correct.

As speed can be changed linearly with time, the reason is false.

Option (c) is correct.

9. In circular motion the acceleration changes regularly where as in projectile motion it is constant. Thus, in circular motion we can't apply v^→=u^→+^→at directly, whereas in projectile motion we can say reason that in circular motion gravity has no role is wrong.

Option (c) is correct.

10. N=mgcos θ

Therefore, assertion is wrong.

Particle performs circular motion due to N sin θ

∴ mv

r N

2

= sin θ

N cos θ is balanced by mg (weight of particle).

Acceleration is not along the surface of the funnel. It is along the centre O of the circle. Thus, reason is true.

Option (d) is correct.

11. Centripetal force mv

r N mg

 2

 

 = +

i.e., Centripetal force (reason ) ≥ wt mg( ) of water

for N = 0, v= gr

If at the top of the circular path v≥ gr i.e., if bucket moved fast, the water will not fall (Assertion).

As assertion and reason both are true and reason is the correct explanation of the assertion of the option would be (a).

Objective Questions (Level 2) Single Correct Option

1. KE 1 2

mv2



 

 = Change in PE 1 2

k x(∆ )2









∆x = Length of spring (Collar at B)

− Length of spring (Collar at A)

= (7+5)² +5²m − 7m = 6 m Thus, 1

2 2 1

2 200 6

2 2

× ×v = × ×

i.e., v²=3600

∴ Normal reaction = mv r

2

=2×3600

5 = 1440 N Option (a) is correct.

2. The particle will remain in equilibrium till ω is constant. Any change in the value of ω will displace the particle up (if ω increase) and down (if ω decreases). Thus, the equilibrium is unstable.

Option (b) is correct.

3. Centripetal force = µ mg or mrω² =µmg or 5

4 a 2

ω =µ g or ω² 4

=15g

a as µ =

 

 1 3 Option (d) is correct.

mg cos θ mg

N O

N sinθ θ

N mg

4. Acceleration at B = Acceleration at A Option (c) is correct.

5. At point P (for the circular motion)

mg N mv

cos θ − = R

2

If at point P skier leaves the hemisphere.

N = 0 Option (c) is correct.

6. Velocity at B v= 2gh

For the circular motion at B, when block just leaves the track

v Option (c) is correct.

7. mg mv Option (c) is correct.

8. v

Option (b) is correct.

10. θ ω= ₀ +1α²

Circular Motion |

153

= + − ⋅

For second hand to meet minute hand for the first time.

2π + Angle moved by minute hand in t second

= Angle moved by second hand in t second or 2 2 Option (d) is correct.

13. (PR)²+(QR)²=(PQ)² (v_A⋅2)² +(v_B⋅2)²=30²

v²_A +v_B² =225 …(i)

Further, PR

PQ=cos 30°

Acceleration of particle when it leaves sphere

= g sin θ

= g 5 3 Option (b) is correct.

H

15. For minimum velocity ( )v : tanθ =tan45° =1

µ = 1 (given)

As µ= tanθ θ is the angle of repose.

Therefore, the automobile will be at the point of slipping when its velocity is zero For maximum velocity ( )v′

N N mv

sinθ+µ cosθ= r′² Also Ncosθ=mg+µNsinθ

i.e., N(cosθ−µsin )θ =mg …(ii) Dividing Eq. (i) by Eq. (ii),

sin cos

cos sin

θ µ θ

θ µ θ

+

− =v′

gr

2

Now, as θ =45 and µ = 1° v′ = ∞ Option (d) is correct.

16. For the particle to just complete the circle the value of

u= 5gR

Particle's velocity ( )v when it is at B i.e., when its velocity is vertical would be given by the relation

v²=u² +2(−g R) =5gR−2gR = 3 gR At B :

a v

R g

c = =

2

3

a_T = g

∴ a_net = a_c²+a²_T

= (3g)²+ g = g 10² Option (a) s correct.

17. N=mrω² and µN=mg

∴ µ

= ωg r ² = 10

5² 0.2( ) = 0.2 Option (c) is correct.

18. v gra

max= h

= 10×200×075. 1.5 = 31.62 m/s

19. At point A : Velocity is zero and as such its vertical component will also be zero.

At point B : Velocity is completely horizontal and as such its vertical component will again be zero.

In figure, TP=Rcos θ At point P :

v²=2g R( cos )θ v= 2gRcos^{1 2/} θ

∴ v_⊥ =vsin θ= 2gR cos^{1 2/} sinθ For v_⊥ to be maximum

d d v

θ ^⊥ = 0 i.e., d

d gR

θ 2 cos^{1 2/} θsinθ=0 or cos^{1 2/} cos 1cos ^{1 2/} ( sin ) sin

2 0

θ θ+ ⁻ θ− θ θ=

or cos cos sin

θ θ cosθ

− θ=

2

2 0

Circular Motion |

155

R

v

u = √5gR

µN N mg

θ

N cos θ N

µN sin θ µN cos θ

µN N sin θ

v1

θ A T

B v

P R

or 2cos²θ−sin²θ=0 or 3cos θ =² 1 or θ =cos⁻¹ 1

3 Option (b) is correct.

20. At any time

Option (c) is correct.

21. At highest point B

T mg mv

If u = velocity at the lowest point A v² =u²+2(−g) ( )2l or gl=u² −4gh

⇒ u² =5gl i.e., u= 5gl (If T = 0, T a^{→ →}⋅ = 0) Option (b) is correct.

22. For any value of u at the lowest point both T→

and a^→ will be towards the centre of the circle and thus

T a→ →

⋅ will be positive.

Option (d) is correct.

23. Change in PE of the system

=Mgl +

24. Decrease in KE = Increase in PE i.e., 1

2 2

mu2 M

= +m gl

 



∴ Initial speed given to ball,

u M m

m gl

=  +

 

 2

25. Maximum and minimum velocities will be respectively at the lowest and the highest points.

v² =u²+2(−g)2L

=u²−4gl

=(2v)²−4gL (Q u= 2 )v

or 3v²=4gL

i.e., v gL gL

= 4 =

3 2

3 Option (b) is correct.

26. u v gL

=2 =4 3

∴ KE at the lowermost position =1 2

mu2

=8 3 mgL

27. Let ω = velocity of the particle when moving downwards.

∴ ω²=v² +2gL = 4 +

3gl 2 gl

or ω = 10

3 gl

Option (a) is correct.

More than One Correct Options 1. At point C :

a_T is maximum and a_c is somewhere between maximum value (at D) and minimum value (at A).

∴ Option (b) is correct.

At point D :

a_c is maximum while be a_T is minimum.

∴ Option (d) is correct.

2. At point B :

T mg mv

+ = l′²

or 2

2

mg mg mv

+ = l′

or v′ =² 3 gl i.e., v′ = 3gl Option (d) is correct.

Now, v²= ′ +v² 2g( )2l

∴ v² =3gl+4gl = 7 gl i.e., v= 7gl Option (b) is correct.

Circular Motion |

157

v

u ω

(Max. velocity) (Min. velocity)

T B v'

v mg

l g

a =T g sin θ B

a = 0T

v_min A

v_max

D a = 0T

g Ca = gT

a (max)c

a (min)c

θ

3. Nsin α =mg

Ncos α=mrω²

⇒ tan α

= ωg

r ² = g r tan θ i.e., 2π

T α

g

= r tan

or T r

= 2π tang α

If α is increase r will also increase and as such T will increase.

Option (c) is correct.

T h

= 2 g

2

π tan α

Thus, if h is increase, T will increase.

Option (a) is correct.

4. Particle can’t have uniform motion because of change in direction of motion i.e., its velocity value. [Option (a)]

Particle can’t have uniformly accelerated motion as acceleration changes direction even if speed is constant. [Option (b)]

Particle can’t have not force equal to zero as centripetal force would be required for the circular motion. [Option (c)]

5. For this see figure in answer 3.

If ω is increased N cos α will increase.

Thus, N will increase (as α is constant) [Option (b)]

And as such net force

= N²+m g^{2 2}+2Ncos (90° +α) on the block will increase.

∴ Option (a) is correct.

As N increases, the value of N sin α (acting opposite to mg) will increase and the block will upwards i.e., h will increase.

Match the Columns 1. At point B :

v²=u² −2gl =12gl−2gl i.e., v= 10gl = 10×10×1 = 10 ms⁻¹ (a) → (p).

Acceleration of bob : a v

l g

c = = =

2 10

1 100 ms⁻² a_T = g= 10 ms⁻² a_net = a²_c +a_T²

= 100²+10² ms⁻² (b) → (s).

Tension in string : T mv

= l = ×

=

2 1 100

1 100 N

(c) → (r).

Tangential acceleration of bob : a_T = g= 10 ms⁻² (d) → (p).

2. v= 2t

∴ dv

dt = 2 i.e., a = 2 ms⁻²

At t = 1 s : v = 2 ms⁻¹

∴ a v

c = r = =

2 22

2 2 ms⁻² a_T = 2 ms⁻²

∴ a_net = 2²+2² =2 2 ms⁻² Angle between a^→net and v^→=45°

[As a_c =a_T] v

B

A b

N sin θ

mg N

h r

α N cos θ

(a) a v^{→ →}⋅ =| || |cos 45^→a v^→ °

=^→anet⋅v 1 2

=2 2 2⋅ 1 2

= 4 unit.

Thus, (a) → (r).

(b) |^→a×^→ω| | || |sin= ^→v ^→ω 90°

[ω^→ will be perpendicular to the plane of circle]

= anetω = a v

net r =2 2⋅ 2

r = 2 2 unit Thus, (b) → (p)

(c) v^{→ →}⋅ω=| || |sin 90^{→ →}v ω ° =v = ⋅v v

ω r =( )2 2

2

= 2 unit Thus, (c) → (q)

(d) |^→v×a^→| | || |sin= ^→v a^→ 45°

=v a_net ⋅ 1 2

= 4 unit Thus, (d) → (r)

3. N M

g v h

1 ra

2

= 2  −

 



On increase v, the value of N₁ will decrease.

Thus, (a) → (q).

N M

g v h

2 ra

2

= 2  +

 



On increasing v, the value of N₂ will increase.

∴ (b) → (q).

As the centripetal force ( )F would be provided by the frictional force (f )

F= f

i.e., F

f = 1

∴ (c) → (r).

If v is increased F will increase which will automatically increase the value of f .

∴ (d) → (p).

4. Speed of particle is constant

∴ | | | |

| |

a v

r

→ →

= → 2

( ) ( ) ( ) ( )

( ) ( )

− + = + −

+ −

6 4

3 4

2 2 2 2

2 2

b a

36 16

5

2 2

+ = +

b a

…(i) (A) r v^{→ →}⋅ =| || |cos 90r v^{→ →} ° =0

i.e., (3^{^}i−4^{^}j) (⋅ 4^{^}i−a^{^}j)=0 12+4a=0 a = − 3

∴ (a) → (s).

(B) Substituting value of a (= −3 in Eq. (i)) 36+b² =5

36+b² =25 b = 25−36

∴ (b) → (s).

(C) r^→=3^{^}i−4^{^}j

∴ r =| |^→r =5

∴ (c) → (r).

(D) (^→v×a^→) | || |sin= ^→v a^→ 90 k°^{^}

=| || |^→v a k ^{→ ^} Thus, ^→r⋅(^→v×^→a)

=^{→ → →}r v a k⋅| || |^{^}

=(3^{^}i−4^{^}j) | || |⋅^→v a k = 0^{→ ^}

[as i k^{^}⋅^{^} = 0, j k^{^}⋅^{^} = 0]

∴ (d) → (s).

Circular Motion |

159

5. (A) As speed is constant Average speed = 1 ms⁻¹

∴ (a) → (s)

(B) Modulus of average velocity = AB t_AB

= R = ⋅

AB

R R

2 2

2 arc 1 speed

π =2 2 π

∴ (b) → (q)

(C) Modulus of average acceleration =| |a^→_c [as speed is constant]

= −

=

→ → →

|v₂ v₁| | v| t_AB t_AB

∆ =v R

2 π2 = 2 [ Q v = 1 ms⁻¹ and R = 2

π]

∴ (c) → (r)

B

A

∆v mg v1

v₂

v1

→

→ → →

Introductory Exercise 8.1

1. If a body is placed in a uniform gravitational field, the CG of the body coincides with the CM of the body.

r

→ = r

→

= CM =

Σ Σ

i n

i i

i n

i

m m

1

1

while r

→ = r

→

= CM=

Σ Σ

i n

i i i

i n

i i

m g m g

1

1

If a body is placed in a uniformly increasing gravitational field (^→g) in the upward direction the CG of the body will be higher level than the CM. And, if the body is placed in a uniformly decreasing gravitational field in the upward direction the CG will be at a lower level the CM.

CG shifts from CM according to the magnitude and direction of the gravitational field (by some other agency eg, earth) in which the body is placed.

In zero gravitational field CG has no meaning while CM still exists, as usual.

2. The cen tre of mass of a rigid body may lie in side, on the sur face and even out side the body. The CM of a solid uni form sphere is at its cen tre. The CM of a solid ring is at the cen tre of the ring which lies out side the mass of the body thus, the state ment is false. (For fur ther de tails see an swer to 1 As ser tion and Rea son JEE corner).

3. Cen tre of mass al ways lies on the axis of sym me try of the body, if it ex ists. The state ment is thus true.

4. State ment is true.

5. As more mass is towards base.

Distance < r 4.

6. If two equal masses are kept at co-ordinates (R, 0) and (−R 0 , then their, ) centre of mass will lie at origin.

7.

X m X m X

m m

CM= +

+

1 1 2 2

1 2

Y m y m y

m m

CM= +

+

1 1 2 2

1 2

r= X_CM² +Y_CM²

= Distance of centre of mass from A g

increasing

CG

CM CG

CM g

decreasing

Centre of Mass, Conservation

In document DC Pandey Mechanics Part 1 Solutions PDF (Page 149-160)