IIT JEE , AIEEE Forum - Physics , Mechanics

harsha19

consider a simple pendulum with inextensible string and solid metal bob.centre of bob is at a height 'H' from the ground. the bbob is given a velocity 'u'due to which it goes to a height 'h' over 'H' . At this instant the string is cut.[The direction of bob makes an angle

with the horizontal when the string is cut.]This bob falls on a spring placed at some distance and copresses it by 'x'.The spring performs simple harmonic motion.The time period of the spring is a] when the bob falls vertically on the spring? b]when the bob falls at an angle

with horizontal on the spring ?

elessar_iitkgp

Lets solve the second case first. The first case is a special case of the solution of the second one.
By energy conservation,
(1/2)mv₀² + mg(h+H) = (1/2)mu² + mgH
v₀² = u² - 2gh ------------(1)
where v₀ is the speed of the bob when the string is cut.

Also,
L(1-cos

) = h
L = h/(1-cos

) -------------(2)
where L is the length of the string

After the string is cut the bob moves as a projectile, projected from a height h+H, at an angle

.

At any time t,
v(t) = v₀cos

i + (v₀sin

- gt) j
Let at t =T the velocity makes an angle of

with the horizontal, and let the speed at this instant be V

tan

= (v₀sin

- gT)/v₀cos

T = (v₀sin

- v₀cos

tan

)/g ------------(3)

Now, when it falls on the spring, the vertical component of velocity is responsible for the compression.
(1/2)m{Vsin

)² = (1/2)kx²
m/k = x²/{Vsin

)² ------------(4)

From (3)
vsin

- gT = vcos

tan

Hence
v(T) = vcos

i + vsin

tan

j
The magnitude of v(T) is V
V =

(v₀²cos²

+ v₀²sin²

tan²

+v₀²sin2

tan

)
V = v₀

(cos²

+ sin²

tan²

+sin2

tan

) ----(5)

= 2

(m/k) = 2

{x/(Vsin

)
=2

{x/

[(u² - 2gh)sin

(cos²

+ sin²

tan²

+sin2

tan

)]}

cvramana

Method suggested by elessar_iitkgp is correct.

elessar_iitkgp

I have modified the solution. Please check the steps

kulki_1123

all i wanted to know is that if there is no restoring force acting on the metal bob after the string is cut(assuming that the cutting does not produce any force).... then the bob should fall in the downward direction and not as a projectile... then the problem cud b made simpler...

elessar_iitkgp

Please check the solution now. Earlier I had taken the wrong angle of projection.

elessar_iitkgp

Solution to the first part

The string is cut when the string is horizontal. Hence L = h, where L is the length of the string. Hence

=

/2

By energy conservation,
(1/2)mv₀² + mg(h+H) = (1/2)mu² + mgH
v₀² = u² - 2gh ------------(1)
where v₀ is the speed of the bob when the string is cut and

After the string is cut the bob moves as a particle projected up from a height h+H.

At any time t,
v(t) = (v₀ - gt) j

Let at t =T the ball falls on the pan.Let its speed be V then.
(1/2)mV² = (1/2)kx²
V = x

(k/m)

Assuming that the bob is moving down when it hits the pan,
gT - v₀ = x

(k/m)
T = (v₀ + x

(k/m))/g

I can't proceed further. Some more information seems to be needed. Or my solution might be incorrect. Please check the post and see if you can suggest anything

harsha19

i gave that the direction of bob makes an angle

with the horizontal.actuallly i have prepared this question .iam not able to find where the mistake is, please help me in correcting the question.

harsha19

please try it once again and reply soon

elessar_iitkgp

I'll try and frame the question for you.

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