Question

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

Answer 1

Answer:

(a)$v_c=1.56\ m.s^{-1}$

(b)$KE=5.3539\ J$

(c-i)$v_h=3.12\ m.s^{-1}$ in straight rightward direction.

(c-ii)$v_l=0\ m.s^{1}$

(c-iii)$v_r=2.2062\ m.s^{-1}$  $\theta=45^{\circ}$ to the bottom of horizontal right.

(d-i) $v_h=1.56\ m.s^{-1}$ to the horizontal right.

(d-ii)$v_l=1.56\ m.s^{-1}$  horizontally left

(d-iii)$v_r=1.56\ m.s^{-1}$  moving vertically downward

Explanation:

Given:

mass of hoop, $m=2.2\ kg$

diameter of hoop, $d=1.2\ m$

angular speed of hoop, $\omega=2.6\ rad.s^{-1}$

So, time taken for 1 revolution(2π radians) of the hoop:

$T=\frac{2\pi}{2.6}\ s$

(a)

The center will move linearly in the right direction.

circumference of the hoop:

$c=\pi.d$

$c=\pi\times 1.2$

Now the speed of center:

$v_c=\frac{c}{T}$

$v_c=\frac{\pi\times 1.2}{(\frac{2\pi}{2.6})}$

$v_c=1.56\ m.s^{-1}$

(b)

Moment of inertia for ring about central axis:

$I=m.r^2$

where 'r' is the radius of the hoop.

$I=2.2\times 0.6^2$

$I=0.792\ kg.m^2$

∴Kinetic energy

$KE=\frac{1}{2} I.\omega^2+\frac{1}{2} m.v_c^2$

$KE=\frac{1}{2} 0.792\times 2.6^2+\frac{1}{2} 2.2\times 1.56^2$

$KE=5.3539\ J$

(c) (i)

The highest point on the hoop will have the maximum velocity.

Given by:

$v_h=\omega \times d$

$v_h=2.6\times 1.2$

$v_h=3.12\ m.s^{-1}$ in straight rightward direction.

(c) (ii)

Lowest point n the hoop will seem stationary for an observer on the ground.

$v_l=0\ m.s^{1}$

(c) (iii)

Velocity of the right-most point of the loop.

This velocity will have 2 components horizontal right and vertical down.

$v_r=\sqrt{v_c^2+v_d^2}$

here: $v_d$ is the downward component.

$v_d=r.\omega$

$v_d=1.56\ m.s^{-1}$

$\therefore v_r=\sqrt{1.56^2+1.56^2}$

$v_r=2.2062\ m.s^{-1}$

$tan\theta=\frac{1.56}{1.56}$

$\theta=45^{\circ}$ to the bottom of horizontal right.

When the observer is moving in the same direction with $v_c$ velocity:

(d) i

Then,

$v_h=\omega \times r$

$v_h=2.6\times 0.6$

$v_h=1.56\ m.s^{-1}$ to the horizontal right.

(d) ii

The bottom point of hoop will seem to move horizontally left with velocity:

$v_l=r.\omega$

$v_l=0.6\times 2.6$

$v_l=1.56\ m.s^{-1}$  horizontally left

(d) iii

Contrary to the case of stationary observer, this observer will see the extreme right point of the hoop moving vertically downward with a velocity:

$v_r=r.\omega$

$v_r=0.6\times 2.6$

$v_r=1.56\ m.s^{-1}$