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MariettaO [177]
3 years ago
7

What is momentum? And it’s types?

Physics
2 answers:
nekit [7.7K]3 years ago
7 0

Answer:

Explanation: (Momentum is mass * velocity. Its units, in the SI system are kilogram meter per second) What is Inertia?

In physics, momentum is the product of mass and velocity. The greater the product of this equation, the greater the momentum. In science, there are two types of momentum: angular and linear, which concern different types of moving objects. However, that is not where the definition stops. The word momentum is also used as part of everyday speech, and is one of the very few physics concept whose metaphorical definition is the same as its actual one.

I HOPE THIS ANSWER WILL HELP YOU!!!

uranmaximum [27]3 years ago
3 0

Answer:

Momentum is the product of mass and velocity. The two types of momentum are angular and linear.

Explanation:

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A uniform rod of mass M and length L can pivot freely at one end. Initially, the rod is oriented vertically above the pivot, in
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Answer:

The speed of its center of mass =\sqrt{\frac{3}{2}gL}

Explanation:

Consider the potential energy at the level of center of mass of rod below the pivot=0

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Length of rod=L

The rotational inertia about the end of a uniform rod=\frac{1}{3}ML^2

Kinetic energy at the level of center of mass of rod below the pivot=\frac{1}{2}I\omega^2

Kinetic energy at the level of center of mass of rod above the pivot=0

Potential energy at the level of center of mass of rod above the pivot=mgh

We have to find the center of mass ( in terms of g and L).

According to conservation of law of energy

Initial P.E+Initial K.E=Final P.E+Final K.E

mgh+0=0+\frac{1}{2} I\omega^2

Where K.E=\frac{1}{2} I\omega^2

I=Moment of inertia

\omega=Angular velocity

Substitute the values then we get

MgL=\frac{1}{2}\times \frac{1}{3}ML^2\omega^2

\omega^2=\frac{6g}{L}

Now, we know that \omega=\frac{v}{r}, r=\frac{L}{2}

Substitute the values then we get

\frac{v^2}{(\frac{L}{2})^2}=\frac{6g}{L}

\frac{v^24}{L^2}=\frac{6g}{L}

v^2=\frac{6g\times L^2}{4L}

v^2=\frac{3gL}{2}

v=\sqrt{\frac{3}{2}gL}

Hence, the speed of its center of mass =\sqrt{\frac{3}{2}gL}

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4 years ago
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