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3 years ago

A cricketer moves his hands backwards when holding a catch

1 answer:

3 0

By doing this he increases the time due to which the rate of change momnetemdecreases. that means the ball comes to rest gently and does not hurt the fielder. but, if does not move his hand backwards, then the ball would come to rest in a fraction of second with a high rate of change of momentum. the ball will exert a lot of force on the hands of the fielder which will hurt him.

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Read 2 more answers

Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas

Zielflug [23.3K]

Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

$I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}$

For 4 particles:

$I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

$I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}$

$I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}$

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