SOLUTIONS TO CONCEPTS

circular motion;;

CHAPTER 7 1.

Distance between Earth & Moon 5 8 r = 3.85 × 10 km = 3.85 × 10 m 6 T = 27.3 days = 24 × 3600 × (27.3) sec = 2.36 × 10 sec v=

2r 2  3.14  3.85  10 8 = = 1025.42m/sec T 2.36  10 6

v2 (1025.42)2 2 –3 2 = = 0.00273m/sec = 2.73 × 10 m/sec r 3.85  10 8 Diameter of earth = 12800km 5 Radius R = 6400km = 64 × 10 m a=

2.

V=

2  3.14  64  10 5 2R = m/sec = 465.185 T 24  3600

V2 ( 46.5185 )2 2 = = 0.0338m/sec R 64  10 5 V = 2t, r = 1cm a) Radial acceleration at t = 1 sec. a=

3.

v2 22 2 = = 4cm/sec r 1 b) Tangential acceleration at t = 1sec. dv d = (2t ) = 2cm/sec2 a= dt dt c) Magnitude of acceleration at t = 1sec

a=

2 4 2  22 = 20 cm/sec Given that m = 150kg, v= 36km/hr = 10m/sec, r = 30m

a=

4.

Horizontal force needed is 5.

in the diagram R cos  = mg

mv 2 150  (10 )2 150  100 = = = 500N r 30 30

..(i)

2

mv ..(ii) r Dividing equation (i) with equation (ii) R sin  =

Tan  =

2

mv2/R

2

mv v = rmg rg

v = 36km/hr = 10m/sec, Tan  =

R

mg

r = 30m

100 v2 = = (1/3) 30  10 rg –1

6.

  = tan (1/3) Radius of Park = r = 10m speed of vehicle = 18km/hr = 5 m/sec

v2 rg v2 25 –1 –1 rg = tan 100 = tan (1/4)

Angle of banking tan =   = tan

–1

7.1

Chapter 7 7.

The road is horizontal (no banking) R

mv 2 = N R and N = mg mv 2 =  mg v = 5m/sec, R 25 25  = g   = = 0.25 10 100 Angle of banking =  = 30° Radius = r = 50m

So

8.

tan  = 

1 3

v= 9.

mv2/R

g

R = 10m mg

v2 v2  tan 30° = rg rg =

rg 50  10 v2 2 v = = rg 3 3 500 3

= 17m/sec.

Electron revolves around the proton in a circle having proton at the centre. Centripetal force is provided by coulomb attraction. –11 –3 r = 5.3 t 10 m m = mass of electron = 9.1 × 10 kg. –19 charge of electron = 1.6 × 10 c.

mv 2 kq2 23.04 q2 9  10 9  1.6  1.6  10 38 2 = k 2 v = = =  1013 r rm 48.23 r 5.3  10 11  9.1 10  31 2 13 12  v = 0.477 × 10 = 4.7 × 10 6

 v = 4.7  1012 = 2.2 × 10 m/sec 10. At the highest point of a vertical circle

mv 2 = mg R 2

 v = Rg  v =

Rg

11. A celling fan has a diameter = 120cm. Radius = r = 60cm = 0/6m Mass of particle on the outer end of a blade is 1g. n = 1500 rev/min = 25 rev/sec  = 2 n = 2  ×25 = 157.14 2 Force of the particle on the blade = Mr = (0.001) × 0.6 × (157.14) = 14.8N The fan runs at a full speed in circular path. This exerts the force on the particle (inertia). The particle also exerts a force of 14.8N on the blade along its surface. 1 12. A mosquito is sitting on an L.P. record disc & rotating on a turn table at 33 rpm. 3 1 100 n = 33 rpm = rps 3 3  60 100 10  = 2  n = 2  × = rad/sec 180 9 r = 10cm =0.1m, g = 10m/sec2

 10  0.1    r 2  9  mg  mr   =  10 g

2

2



2  81 7.2

Chapter 7 13. A pendulum is suspended from the ceiling of a car taking a turn 2 r = 10m, v = 36km/hr = 10 m/sec, g = 10m/sec

mv 2 T sin  = r T cos  = mg

From the figure



 mv2/R

..(i) mg

..(ii)

2 sin  mv 2 v2 –1  v  =  tan  =   = tan    rg  cos  rmg rg   100 –1 –1 = tan 10  10 = tan (1)   = 45°

14. At the lowest pt. T

mv 2 r Here m = 100g = 1/10 kg,

T = mg +

mv T = mg + r

2

(1.4) 1 =  9 .8  10 10

r = 1m,

v = 1.4 m/sec

mg

2

= 0.98 + 0.196 = 1.176 = 1.2 N

mv2 /r

15. Bob has a velocity 1.4m/sec, when the string makes an angle of 0.2 radian. m = 100g = 0.1kg, r = 1m, v = 1.4m/sec. From the diagram,

mv R

T – mg cos  =

mg sin 

mv 2 T= + mg cos  R T=

T

2

mg cos 

 2  0.1 (1.4)2   (0.1)  9.8  1  1 2  

 (.2)2    T = 0.196 + 9.8 × 1   2  

( cos  = 1 

2 for small ) 2

 T = 0.196 + (0.98) × (0.98) = 0.196 + 0.964 = 1.156N  1.16 N 16. At the extreme position, velocity of the pendulum is zero. So there is no centrifugal force. So T = mg cos o 17. a) Net force on the spring balance. 2 R = mg – m r So, fraction less than the true weight (3mg) is

T

mg sin 

mg cos 

2

2 mg  (mg  m2r ) 2   6400  10 3 –3 = =  = 3.5 × 10   mg g 10  24  3600 

=

b) When the balance reading is half the true weight, 2

mg  (mg  m r ) = 1/2 mg 2

 r = g/2  

g 10  rad/sec 2r 2  6400  10 3

 Duration of the day is T=

2 2  8000 2  6400  10 3 64  10 6 = 2  sec = 2  sec = hr = 2hr  7  3600 9 .8 49

7.3

R mg 2

m /R

Chapter 7 18. Given, v = 36km/hr = 10m/s, r = 20m,  = 0.4 The road is banked with an angle,  v2   1  –1  –1  100  –1  = tan  rg  = tan   = tan  2  or tan  = 0.5    20 10  When the car travels at max. speed so that it slips upward, R1 acts downward as shown in Fig.1 So, R1 – mg cos  –

tan    = rg 1   tan 

V1 =

 

mv1 /r

R1

mg

..(i)

R2

2

mv 1 cos  = 0 r Solving the equation we get,

And R1 + mg sin  –

R1

2

mv 1 sin  = 0 r

2



2

mv2 /r

..(ii) 

0 .1 = 4.082 m/s = 14.7 km/hr 20  10  1 .2



So, the possible speeds are between 14.7 km/hr and 54km/hr. 19. R = radius of the bridge L = total length of the over bridge a) At the highest pt.

mv 2 2  v = Rg  v = R 1 b) Given, v = Rg 2

mg =

R2

mg

mv2/R

Rg mg 2

2= L/R mv2/R

mv 2 suppose it loses contact at B. So, at B, mg cos  = R 2  v = Rg cos  2  Rv   Rg   = Rg cos   cos  = 1/2   = 60° = /3  2  = Rg cos     2

 mg 2 2= L/R

 R  =  ℓ = r = r 3 R from highest point 3 c) Let the uniform speed on the bridge be v. So, it will lose contact at distance

The chances of losing contact is maximum at the end of the bridge for which  =

mv2/R

L . 2R

mv 2 = mg cos   v = R

 L  gR cos   2R  20. Since the motion is nonuniform, the acceleration has both radial & tangential component So,

 2

2= L/R

m

2

v r dv at = =a dt

ar =

 mg

mv2/R

mv2/R

2

Resultant magnitude =

 v2     a2  r   

m dv/dt

2

 v2  Now N = m    a 2   mg = m  r   4

2 2

2

2

2 2

2

2

 v2     a 2  2g2 =  r    2 1/4

 v = ( g – a ) r  v = [( g – a ) r ]

7.4

 v4  2    r2   a  

m

N

Chapter 7 21. a) When the ruler makes uniform circular motion in the horizontal plane, (fig–a)  mg = mL

mg 

g  L b) When the ruler makes uniformly accelerated circular motion,(fig–b) 

 mg =

2

2

(m2 L )  (mL  )

2



4 2

 2 g2 + =  2 = L2 2

 g  2      2   L  

 mg

12L

L 1/ 4

R

(Fig–a) m22L

 mg

mL

(When viewed from top)

(Fig–b)

22. Radius of the curves = 100m Weight = 100kg Velocity = 18km/hr = 5m/sec a) at B mg –

mv 2 100  25 = N  N = (100 × 10) – = 1000 – 25 = 975N R 100

mv 2 = 1000 + 25 = 1025 N A R b) At B & D the cycle has no tendency to slide. So at B & D, frictional force is zero. 1 = 707N At ‘C’, mg sin  = F  F = 1000 × 2 At d, N = mg +

 B

C

E D

mv2/R B

mv 2 mv 2 c) (i) Before ‘C’ mg cos  – N =  N = mg cos  – = 707 – 25 = 683N R R

mg N

mv 2 mv 2 N= + mg cos  = 25 + 707 = 732N R R d) To find out the minimum desired coeff. of friction, we have to consider a point just before C. (where N is minimum) Now,  N = mg sin    × 682 = 707 So,  = 1.037 F m2r 23. d = 3m  R = 1.5m R = distance from the centre to one of the kids N = 20 rev per min = 20/60 = 1/3 rev per sec = 2r = 2/3 15kg 15kg m = 15kg (ii) N – mg cos  =

mg ( 2 ) 2 2 2 = 5 × (0.5) × 4 = 10 9 2  Frictional force on one of the kids is 10  24. If the bowl rotates at maximum angular speed, the block tends to slip upwards. So, the frictional force acts downward. Here, r = R sin  From FBD –1 2 R1 – mg cos  – m (R sin ) sin  = 0 ..(i) [because r = R sin ] 2 and R1 mg sin  – m1 (R sin ) cos  = 0 ..(ii) Substituting the value of R1 from Eq (i) in Eq(ii), it can be found out that 2

 Frictional force F = mr = 15 × (1.5) ×

1/ 2

 g(sin    cos )  1 =    R sin (cos    sin )  Again, for minimum speed, the frictional force R2 acts upward. From FBD–2, it can be proved R1 that,

R2

m12 r  (FBD – 1)

7.5

R1

m22 r  (FBD – 2)

R2

Chapter 7 1/ 2

 g(sin    cos )  2 =    R sin (cos    sin )   the range of speed is between 1 and 2 25. Particle is projected with speed ‘u’ at an angle . At the highest pt. the vertical component of velocity is ‘0’ So, at that point, velocity = u cos  mv2/r u sin   2 2 centripetal force = m u cos  r  u cos  At highest pt.

mg

mv 2 u2 cos 2  mg = r=  r g 26. Let ‘u’ the velocity at the pt where it makes an angle /2 with horizontal. The horizontal component remains unchanged mv2/ u cos  ...(i) So, v cos /2 =  cos   v =   cos  2 mg  From figure  cos

/ mgcos/2

mv 2 v2 r= r g cos / 2 putting the value of ‘v’ from equn(i)

mg cos (/2) =

u 2 cos 2   g cos 3 ( / 2) 27. A block of mass ‘m’ moves on a horizontal circle against the wall of a cylindrical room of radius ‘R’ Friction coefficient between wall & the block is . r=

a) Normal reaction by the wall on the block is = b)  Frictional force by wall =

mv 2 R

mv 2 R 

mv 2 v 2 c) = ma  a = – (Deceleration) R R 2

dv dv v = v =– dt ds R

d) Now, s= 

R

 ds = –

m

R

R dv  v

In V + c

mv2/R

At s = 0, v = v0 Therefore, c = so, s = 

R In V0 

mv 2/R

R v v –s/R In  =e  v0 v0 –2

For, one rotation s = 2R, so v = v0e 28. The cabin rotates with angular velocity  & radius R 2  The particle experiences a force mR . 2 The component of mR along the groove provides the required force to the particle to move along AB. 2 2  mR cos  = ma  a = R cos  B length of groove = L A  2 2 2 L = ut + ½ at  L = ½ R cos  t 2 2

t =

2L 2L =t= 1  R2 cos  R2 cos 

R

7.6

mv /R

Chapter 7 29. v = Velocity of car = 36km/hr = 10 m/s r = Radius of circular path = 50m m = mass of small body = 100g = 0.1kg.  = Friction coefficient between plate & body = 0.58 a) The normal contact force exerted by the plate on the block

mv 2 0.1 100 = = 0.2N r 50 b) The plate is turned so the angle between the normal to the plate & the radius of the road slowly increases N=

N=

mv 2 cos  r

..(i)

mv 2 sin  ..(ii) r Putting value of N from (i)

N=

mv 2 mv 2 –1 –1 cos  = sin    = tan    = tan  = tan (0.58) = 30°  r r 30. Let the bigger mass accelerates towards right with ‘a’. From the free body diagrams, …(i) T – ma – mR = 0 2 T + 2ma – 2m R = 0 …(ii) 2 Eq (i) – Eq (ii)  3ma = m R 

m2R 3 2 Substituting the value of a in Equation (i), we get T = 4/3 m R. a=

****

7.7

m 2m

R

ma

T

2ma

T

a

m2R 2m2R