What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise

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The bob of a pendulum is released from a horizontal position A as shown in the figure. If the length of the pendulum is 2 m, what is the speed with which the bob arrives at the lowermost point B, given that it dissipated 10% of its initial potential energy w.r.t. B point against air resistance? (g=10 m s-2)

A 1 kg block situated on a rough inclined plane is connected to a spring of spring constant 100 N m–1 as shown in the figure. The block is released from rest with the spring in the unstretched position. The block moves 10 cm along the incline before coming to rest. Find the coefficient of friction between the block and the incline assume that the spring has negligible mass and the pulley is frictionless. Take g=10 m s–2.

1. Work done:

W=∫dW=∫F→⋅dr→=∫Fdrcosθ,

[where θ is the angle between F→ & dr→]

(i) For constant force W=F→⋅d→=Fdcosθ

(ii) Calculation of work done from force-displacement graph:

W=∫dW=∫Fdx= Area between F- x curve and x-axis

(iii) Nature of work done: Although work done is a scalar quantity, yet its value may be positive, negative or even zero

2. Work done by multiple forces:

ΣF→=F→1+F→2+F→3+…

W=ΣF→⋅S …i

W=F1⋅S→+F→2⋅S→+F→3⋅S→+…

or W=W1+W2+W3+…

3. Relation between momentum and kinetic energy:

K=p22m and P=2mK; P= linear momentum

4. Potential Energy

∫U1U2dU=-∫r1r2F→⋅dr→ i.e. U2-U1=-∫r1r2F→.dr→=-W

U=-∫∞rF→⋅dr→=-W

5. Conservative Forces

F=-∂U∂r

6. Work-energy theorem

WC+WNC+WPS=ΔK

Modified Form of Work-Energy Theorem

WC=-ΔU

WNC+WPS=ΔK+ΔU

WNC+WPS=ΔE

7. Power:

The average power ( P¯ or pavg) delivered by an agent is given by P¯ or pav=Wt

Instantaneous Power: P=dWdt=F→.dS→dt=F→.dS→dt=F→⋅v→

8. Potential energy curve and equilibrium:

It is a curve which shows change in potential energy with position of a particle.

(i) Stable Equilibrium:

When a particle is slightly displaced from equilibrium position and it tends to come back towards equilibrium then it is said to be in stable equilibrium.

At point C: slope dUdx is negative. So, F is positive.

At point D: slope dUdx is positive. So, F is negative.

At point C: It is the point of stable equilibrium.

U=Umin,  dUdx=0 and d2Udx2= positive

(ii) Unstable equilibrium:

When a particle is slightly displaced from equilibrium and it tends to move away from equilibrium position then it is said to be in unstable equilibrium.

At point E: slope dUdx is positive. So, F is negative.

At point G: slope dUdx is negative. So, F is positive.

At point B: It is the point of unstable equilibrium.

U=Umax,  dUdx=0 and d2Udx2= negative

(iii) Neutral equilibrium:

When a particle is slightly displaced from equilibrium position and no force acts on it then equilibrium is said to be neutral equilibrium. Point H is at neutral equilibrium.

⇒U= constant, dUdx=0,  d2Udx2=0

9. Collision of bodies

The event or the process, in which two bodies either coming in contact with each other or due to mutual interaction at distance apart, affect each other's motion (velocity, momentum, energy or direction of motion) is defined as a collision.

In collision, the particles come closer before collision and after collision they either stick together or move away from each other. The particles need not come in contact with each other for a collision.

The law of conservation of linear momentum is necessarily applicable in a collision, whereas the law of conservation of mechanical energy is not.

10. Coefficient of restitution e:

e=Impulse of reformation  Impulse of deformation =∫Frdt∫Fddt

= Velocity of separation along line of impact  Velocity of approach along line of impact 

11. Head on collision:

Consider the Head on inelastic collision of two particles,

Let the coefficient of restitution for collision is e

(i) Momentum is conserved m1u1+m2u2=m1v1+m2v2 …i

(ii) Kinetic energy is not conserved.

(iii) According to Newton's law e=v2-v1u1-u2 …ii

By solving equations (i) and (ii):

v1=m1-em2m1+m2u1+(1+e)m2m1+m2u2=m1u1+m2u2-m2eu1-u2m1+m2

v2=m2-em1m1+m2u2+(1+e)m1m1+m2u1=m1u1+m2u2-m1eu2-u1m1+m2

12. Elastic Collision e=1

(i) If the two bodies are of equal masses: m1=m2=m,then,  v1=u2 and v2=u1

Thus, if two bodies of equal masses undergo elastic collision in one dimension, then after the collision, the bodies will exchange their velocities.

(ii) If the mass of a body is negligible as compared to other.

(a) If m1>>m2 and u2=0 then v1=u1,v2=2u1

when a heavy body A collides against a light body B at rest, the body A should keep on moving with same velocity and the body B will move with velocity double that of A.

(b) If m2>>m1 and u2=0 then v2=0,v1=-u1

When light body A collides against a heavy body B at rest, the body A should start moving with same velocity just in opposite direction while the body B should practically remains at rest.

13. Inelastic Collision:

Loss in kinetic energy in inelastic collision

ΔK=m1 m22 m1+m21-e2u1-u22

14. Oblique Collision:

Conserving the momentum of system in directions along normal (x axis in our case) and tangential (y axis in our case)

m1u1cosα1+m2u2cosα2=m1v1cosβ1+m2v2cosβ2 and

m2u2sinα2-m1u1sinα1=m2v2sinβ2-m1v1sinβ1

Since no force is acting on m1 and m2 along the tangent (i.e. y-axis) the individual momentum of m1 and m2 remains conserved.

m1u1sinα1=m1v1sinβ1 & m2u2sinα2=m2v2sinβ2

By using Newton's experimental law along the line of impact

e=v2cosβ2-v1cosβ1u1cosα1-u2cosα2

15. Variable mass system:

If a mass is added or ejected from a system, at rate μ kg s-1 and relative velocity v→rel (w.r.t. the system), then the force exerted by this mass on the system has magnitude μv→rel.

Thrust Force F→t:

F→t=v→reldmdt

16. Rocket propulsion:

At t=0 v=u m=m0

At t=t v=v m=m

exhaust velocity =vr

Thrust force on the rocket =vr-dmdt

Velocity of rocket at any instant v=u-gt+vrlnm0m

If gravity is ignored and initial velocity of the rocket u=0,

v=vrlnm0m

17. Circular motion in vertical plane:

(i) Condition to complete vertical circle u ≥ 5gR

(ii) If u=5gR then Tension at C is equal to 0 and tension at A is equal to 6mg.

Velocity at B: vB=3gR

Velocity at C: vC=gR

From A to B:

T=mgcosθ+mv2R

From B to C:

T=mv2R-mgcosθ

(iii) Condition for pendulum motion (oscillating condition)

u ≤ 2gR (in between A to B)

Velocity can be zero but T never be zero between A & B.

Because T is given by T=mgcosθ+mv2R

(iv) Condition for leaving path: 2gR<u<5gR

Particle crosses the point B but not complete the vertical circle

Tension will be zero in between B to C & the angle where T=0

cosθ=u2-2 gR3gR, θ is from vertical line

Note: After leaving the circle, the particle will follow a parabolic path.

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