Question

At time t= 0, a $2.0kg$particle has the position vector$\overrightarrow{r}=\left(4.0m\right)\hat{i}-\left(2.0m\right)\hat{j}$relative to the origin. Its velocity is given by$\overrightarrow{v=}\left(-6.0t2m/s\right)$$\hat{i}$for$t⩾0$in seconds. About the origin, (a) What are the particle’s angular momentum$\overrightarrow{L}$?(b)What are the torque$\overrightarrow{\tau }$ acting on the particle, both in unit-vector notation and for$t>0$? (c) About the point$\left(-2.0m,-3.0m,0\right),$what are$\overrightarrow{L}$?(d) About the pointwhat are$\overrightarrow{\tau }$ for t0?

Short answer

1. The body’s rotational inertia about the rotational axis through its center of mass is$I=\frac{1}{2}m{R}^2$.
2. The body could be a solid cylinder.

Step-by-step solution

Step 1: Given

Body of radius is$R$ ,

Mass is$M$

Speed of the body isrole="math" localid="1663218026097" $v$ .

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:

$K=\frac{1}{2}m{v}_{com}^2$

$K=\frac{1}{2}I{\omega }^2$

$U=Mgh$

where,$\omega$is angular frequency, g is an acceleration due to gravity, h is height, Iis moment of inertia, m, M are masses,$v_{com}^{}$is velocity of centre of mass, U is potential energy and K is kinetic energy.

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

Now,

$K_i=U_f$

$\frac{1}{2}m{v}_{com}^2+\frac{1}{2}I{\omega }^2=Mgh$

$\frac{1}{2}m{v}^2+\frac{1}{2}I{\left(\frac{v}{R}\right)}^2=mg\left(\frac{3V^2}{4g}\right)$

$I=\frac{1}{2}m{R}^2$

Hence,the body’s rotational inertia about the rotational axis through its center of mass is $I=\frac{1}{2}m{R}^2$.

(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.