Mechanics - Kinematics-12

1133

Sourish Ghosh ·2013-06-11 03:15:42

a) v = e^-ktm

b) v = mgk(1 - Ï_lÏ_s)

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Akash Anand do we really need rho-l and rho-s. Can't we do this sum without using that ?? Thinkkkk
Upvote·0· Reply ·2013-06-11 06:31:55
Sourish Ghosh Is the first part correct? Do we have to find v as function of time?
Upvote·0· Reply ·2013-06-11 07:01:50
Akash Anand Yeah you have to ...according to the condition provided
Upvote·0· Reply ·2013-06-12 02:06:41
Sourish Ghosh Any hints?
Upvote·0· Reply ·2013-06-12 03:56:40
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1133

Sourish Ghosh ·2013-06-12 05:50:41

Alright here's what I have done. Where am i going wrong?

When body attains terminal vel (v_t), net force is zero.

F = -kv
=> mdvdt = -kv
on solving,
v = e^-kt/m
a = -kme^-kt/m

mg = F_b + kv_t
=> F_b = mg - kv_t

At any general point before terminal velocity,

mg + ma = F_b + kv
=> ma = k(v - v_t)

on substituting values,
v_t = 2e^-kt/m

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Akash Anand You are wrong at putting the limits of first integration, you should get a constant and then you gonna get the value of that constant. That will be the expression for velocity.
Upvote·0· Reply ·2013-06-12 06:25:20
Sourish Ghosh But, at t = 0, v = 0
Upvote·0· Reply ·2013-06-12 07:37:28
Akash Anand Just try to put the limits properly once again ..whatever you are saying is correct.
Upvote·0· Reply ·2013-06-12 23:31:54
Sourish Ghosh ok if i put limits then v/u = e^(t-t0)/k but u is zero :/
Upvote·0· Reply ·2013-06-13 08:09:05
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1161

Akash Anand ·2013-06-14 02:18:10

Here is the solution

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Sourish Ghosh Is F = -kv, the buoyant force?
Upvote·0· Reply ·2013-06-14 03:07:11
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Kinematics-12