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Laws of Motion

Question
CBSEENPH11020092
Wired Faculty App

A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for straight omega space less-than or slanted equal to space square root of straight g over straight R end root. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = square root of fraction numerator 2 straight g over denominator straight R end fraction end root? Neglect friction.

Solution

Let the radius vector joining the bead with the centre make an angle θ, with the vertical downward direction.

             

Here,
OP = R = Radius of the circle
  N = Normal reaction
The respective vertical and horizontal equations of forces can be written as, 
mg = N Cosθ                         ... (i) 

mlω2 = N Sinθ                       … (ii)
In ΔOPQ, we have
Sin θ = l / R 
l = R Sinθ                              … (iii)

Substituting equation (iii) in equation (ii), we get
 
m(R Sinθ) ω2 = N Sinθ 

mR ω2 = N                            ... (iv) 

Substituting equation (iv) in equation (i), we get
mg = mR ω2 Cosθ 

Cosθ = g / Rω2                        ...(v) 

Since cosθ ≤ 1, the bead will remain at its lowermost point for g / Rω2 ≤ 1.
i.e., for                      ω ≤ (g / R)1/2  
For ω = (2g / R)1/2 
rightwards double arrow      ω2 = 2g / R                  ...(vi) 
On equating equations (v) and (vi), we get
  fraction numerator 2 straight g over denominator straight R end fraction = g / RCos θ  
 Cos θ = 1 / 2 
∴ θ = Cos-1(0.5)  =  600 , is the angle made by the radius vector joining the centre to the bead with the vertical downward direction. 

Question
CBSEENPH11020262
Wired Faculty App

Given in the figure are two blocks A and B of weight 20 N and 100 N respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is

  • 100N

  • 80 N

  • 120 N

  • 150 N

Solution

C.

120 N

In the vertical direction, weight are balanced by frictional forces.
As the blocks are in equilibrium balance forces are in horizontal and vertical direction.
For the system of blocks (A+B)
F = N
For block A, fA = 20 N and for block B.
fB = fA +100 = 120 N

Question
CBSEENPH11020263
Wired Faculty App

A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

  • 44%

  • 50%

  • 56%

  • 62%

Solution

C.

56%

Question
CBSEENPH11020270
Wired Faculty App

A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

  • 2g/3

  • g/2

  • 5g/6

  • g

Solution

B.

g/2

For the mass m,
mg-T = ma

As we know, a = Rα    ... (i)
So, mg-T = mRα
Torque about centre of pully
T x R = mR2α ...... (ii)
From Eqs. (i) and (ii), we get 
a = g/2
Hence, the acceleration of the mass of a body fall is g/2.

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