Question

give proper answer........

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Answer 1

In fig. (a), a particle having linear velocity v1 = v(t) for a time t is moving along a circular path of radius r1 = r(t). After a time ∆t, the particle attains a linear velocity v2 = v(t + ∆t) which is same in magnitude as that of v1 but different in direction.

Here the particle sweeps out an angular displacement Δθ during the time ∆t.

Here the radius at time t + ∆t is r2 = r(t + ∆t) which is same in magnitude but different in direction, like the velocity.

Then,

∆r = r(t + ∆t) - r(t)

∆r / ∆t = (r(t + ∆t) - r(t)) / ∆t

For a small change in time ∆t → 0,

lim_(∆t → 0) ∆r / ∆t = dr / dt

lim_(∆t → 0) ∆r / ∆t = v, velocity, → (1)

which is equal to v1 and v2 in magnitude.

In mathematics it is true that the angle between a chord in a circle and a tangent of the circle passing through one of the ends of the chord is half of the central angle made by the chord in the circle. So, since v1 is acting tangentially to the circular path, we can say that angle between v1 and ∆r is ∆θ / 2.

Similarly, if v2 is extended backwards, then the actual angle between v2 and ∆r is ∆θ / 2, because the obtuse angle between v2 and ∆r is π - (∆θ / 2).

Thus, if v1 and v2 are alone combined without changing the direction, the angle will be (Δθ / 2) + (Δθ / 2) = Δθ.

In fig. (b), the velocities v1 and v2 are alone combined with difference of both velocities, ∆v.

Then,

∆v = v(t + ∆t) - v(t)

∆v / ∆t = (v(t + ∆t) - v(t)) / ∆t

For a small change in time ∆t → 0,

lim_(∆t → 0) ∆v / ∆t = dv / dt

lim_(∆t → 0) ∆v / ∆t = a, acceleration. → (2)

Now the triangles formed by the radius vectors along with ∆r and velocity vectors along with ∆v are similar, because the two sides of each triangle are proportional (by the ratio 1 : 1) and the angle between them are same, i.e., ∆θ. Thus, we can have,

∆r / r1 = ∆v / v1

∆v = v ∆r / r

Dividing both sides by ∆t,

∆v / ∆t = (v ∆r) / (r ∆t)

∆v / ∆t = (v / r) (∆r / ∆t)

For a small time ∆t → 0,

lim_(∆t → 0) ∆v / ∆t = (v / r) · [lim_(∆t → 0) (∆r / ∆t)]

From (1) and (2),

a = v² / r

#answerwithquality

#BAL

Answer 2

sonu.

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bhot jyada had se bi jyada .o(〒﹏〒)o

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or to or

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kya ap in rules oo follows kr skte ho???

pr me to puchna bhool gi ki apko isme interst bi h ya ni ???

me bi na shayad ap meko bhool jaoge :(

ok

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