Q86P

Question

At time = 0, a 2.0 kgparticle has the position vector r= 4.0 mi^-2.0 mj^ relative to the origin. Its velocity is given byv=-6.0t2 m/s i^for t 0  in seconds. About the origin, (a) What are the particle’s angular momentum L ? (b)What are the torque τ acting on the particle, both in unit-vector notation and for t> 0 ?  (c) About the point -2.0 m, -3.0 m, 0, what are L ? (d) About the point what are τ for t 0?

Step-by-Step Solution

Verified
Answer
  1. The body’s rotational inertia about the rotational axis through its center of mass is I=12mR2.
  2. The body could be a solid cylinder.
1Step 1: Given

Body of radius is R ,

Mass is M

Speed of the body is  v

2Step 2: Determining the concept

Using the formula for mechanical energy conservation , find the body’s rotational inertia about the rotational axis through its center of mass. According to conservation of energy, energy can neither be created, nor be destroyed.

 

Formulae are as follow:

 K= 12mvcom2

K=12Iω2

U = Mgh


where, ω is angular frequency, g is an acceleration due to gravity, h is height, is moment of inertia, m, M are masses,  vcomis velocity of centre of mass, U is potential energy and K is kinetic energy.

3Step 3: (a) Determining the body’s rotational inertia about the rotational axis through its center of mass

Now,

 Ki=Uf

12mvcom2+12Iω2=Mgh

12mv2+12IvR2=mg3V24g


I=12mR2

 

Hence, the body’s rotational inertia about the rotational axis through its center of mass is I=12mR2.

4Step 4: (b) Determining the shape of the body

 

From rotational inertia, it can be seen that the body could be a solid cylinder.

 

Hence, the body could be a solid cylinder.

 

Therefore, using the conservation of energy, the rotational inertia of the given object can be found. It can be checked that which object has the same formula for the rotational inertia and it can be decided the type of object.