Q 1 :    

A body of m kg slides from rest along the curve of a vertical circle from point A to B in a frictionless path. The velocity of the body at B is ______.

(given, R=14 m,  g=10 m/s2 and 2=1.4 )       [2024]

  • 21.9 m/s

     

  • 10.6 m/s

     

  • 19.8 m/s

     

  • 16.7 m/s

     

(1)

By conservation of mechanical energy, decrease in P.E. = increase in K.E.

mg(R+R2)=12mvB2-02gR(1+12)=vB2

vB2=2×10×14×(1+11.4)

vB=20×24=43021.9 m/s



Q 2 :    

A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle (θ) of thread deflection in the extreme position will be ______.                   [2024]

  • tan-1(2)

     

  • 2tan-1(12)

     

  • tan-1(12)

     

  • 2tan-1(15)

     

(2)

Loss in kinetic energy = Gain in potential energy

12mv2=mg(1-cosθ)

v2=2g(1-cosθ)

Acceleration at lowest point =v2

Acceleration at extreme point =gsinθ

 v2=gsinθ

 sinθ=2(1-cosθ)

tanθ2=12θ=2tan-1(12)



Q 3 :    

A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m, making 10 rpm. The tension in the string when the stone is at the lowest point is (if π2 = 9.8 and g = 9.8 m/s2)               [2024]

  • 17.8 N

     

  • 8.82 N

     

  • 97 N

     

  • 9.8 N

     

(4)

Given that, m=900 gm=9001000 kg=910 kg

r=1 m

ω=2πN60=2π(10)60=π3 rad/sec

T-mg=mrω2    T=mg+mrω2

T=910×9.8+910×1(π3)2

           =8.82+910×π29=9.80 N



Q 4 :    

A bob of mass 'm' is suspended by a light string of length 'L'. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes a half-circle, reaching the topmost position B. The ratio of kinetic energies (K.E)A(K.E)B is ______.            [2024]

  • 3 : 2

     

  • 5 : 1

     

  • 2 : 5 

     

  • 1 : 5

     

(2)

Apply energy conservation

12mVL2=12mVH2+mg(2L)

 VL=5gL

So, VH=gL

(K.E)A(K.E)B=12m(5gL)212m(gL)2=51



Q 5 :    

A bob of mass m is suspended at a point O by a light string of length l and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity v0 at the point 'A'. The string becomes slack when, the bob reaches at the point 'D'. The ratio of the kinetic energy of the bob at the points B and C is          [2025]

  • 2

     

  • 1

     

  • 4

     

  • 3

     

(1)

Applying conservation of mechanical energy,

12mvA2=12mvB2+mgh

 12m(5gl)=12mvB2+mgl2

 5mgl2mgl2=KEB

 KEB=2mgl

         12mvC2=12mvD2+mgl2

 12mgl+mgl2=mgl

          KEC=mgl

 KEBKEc=2



Q 6 :    

A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is:         [2025]

(Take acceleration due to gravity as 10 m/s2)

  • 2+33

     

  • 2+23

     

  • 3+32

     

  • 322

     

(3)

12m×100+0=12mVB2+mg(RR32)

100=VB2+2gR(132)

VB2=10020(23)

 VB2=60+203

K.E.B=12mVB2=m2(60+203)                                   ...(i)

12m(100)=12mVC2+mg(3R2)

100=VC2+60

VC2=40

K.E.C=12mVC2=12m(40)

 KBKC=(VBVC)2=3+32



Q 7 :    

A body of mass 'm' connected to a massless and unstretchable string goes in verticle circle of radius 'R' under gravity g. The other end of the string is fixed at the center of circle. If velocity at top of circular path is ngR, where n1, then ratio of kinetic energy of the body at bottom to that at top of the circle is          [2025]

  • nn+4

     

  • n+4n

     

  • n2n2+4

     

  • n2+4n2

     

(4)

v0=v2+2g(2R)

v0=n2gR+4gR

 kbottomktop=v02v2=n2+4n2