2018-10-01_Physics_For_You

dm

m

l

= dl

0

The moment of inertia of the elementary mass is

given as

dI = (dm)r^2

The moment of inertia of the rod,

I = ³dI Ÿ I = ³r^2 dm

Substituting r = l sin q and dm =

m

l

dl

0

. we obtain

Il

m

l

=∫(s^22 in ) dl

0

θ =m

l

ldl

ml

l

sin l

sin

2

0

(^20)
3
0
2
(^03)
θ (^0) θ=
∫

Ÿ I

ml

=^0

22

3

sin θ

13. (a) : Acceleration of the rod a

F

m

=

Angular acceleration of the rod

α

τ

== =

I

Fl

ml

F

ml

/

12

6

(^2) /
Net acceleration of point P,
aP = a – Dx = 0
Ÿ

F

m

F

ml

xx

l

−=⇒==

6

0

6

1m

14. (d) : The total KE is K = Krod + Khollow (^) shere

• Ksolid (^) sphere
  where Krod =

1

2

mv^2 ...(i)

Since the CM of each sphere moves with a velocity

v

v

CM= 2 ,

Kmv

K

hollowsphereCM R

=+









1

2

(^21)
2
2

=







 +







=

1

22

1 2

3

5

24

(^22)
m vmv ... (ii)
Km
v
solidsphere=    + = mv

1

22

1

2

5

7

40

2

(^2) ...(iii)
Adding equation (i), (ii) and (iii)
km=+vmvm+=vmv

1

2

5

24

7

40

53

60

2222

15. (b) : The angular momentum of the disc about O is

C v

x

y

(x,R)

O

r

C

F

D a

Dx P a

x

  

LmOC=×rvCC+ωI

=+×+







mxiRjvimR 

v

R

() k

1

2

2 3

=+mxiRjv×+imR  v  ×

R

()^1 km vR()ji

2

2 3

+−+=

3

2

3

22

mvRk mvRk mvRk

mvR

 k

16. (c) : As cotton reel
    rolls point of contact of
    reel with ground will
    act as instantaneous
    axis of rotation.
    vP = Z(R + r)

Ÿ (^) ω=

• 
  

= −

v

Rr

P 20 rad s^1

Velocity of centre of reel

vR

vR

C Rr

==p

+

=

×

+

ω = −

()()

620

20 10

4 m s^1

17. (d) : Let aer collision velocity of rod be v and
    angular velocity be Z then

mv 0 = mv + MV Ÿ mv 0 x = mvx +

Ml^2

12

ω

mv 0 = mv + Mxωη

2

12

From above equations,

Mx

MV xV

ωη

ωη

(^22)
12

=⇒= 12

Velocity of father end of the rod = V−lω

2

=−=− 

ωηωηωη^2 η

12 226

1

xxx

If this velocity is opposite to v 0 , then

η

η

6

−< 10 ⇒< 6

18. (d) : The fall of centre of
    gravity h is given by
    L
    h

L

2

2

60

−

















=°cos ,

or h

L

=− °

2

(c 16 os 0 )

vP=6 m s–

vC

L/

L

2

• h

G

h

f

r

L/

60° G¢