Question

A thin uniform straight rod of mass
2 kg and length 1 m is free to rotate
about its upper end when at rest. It
receives an impulsive blow of 10 Ns
at its lowest point, normal to its
length as shown in figure. The
kinetic energy of rod just after
impact is
10 NS​

Answer 1

Answer:

KE=75j

Explanation:

Angular impulse = change in angular momentum

10×1=Iω−0

I= moment of inertia of rod about it upper end =

12

1

mL

2

+m(

2

L

)

2

=

3

1

mL

2

3

1

mL

2

ω=10

kinetic energy will be(KE) =

2

1

Iw

2

KE=

2

1

×(

3

1

mL

2

)×(

mL

2

30

)

2

KE=

2mL

2

300

KE=

2

150

KE=75J