Answer:
(a)![v_c=1.56\ m.s^{-1}](https://tex.z-dn.net/?f=v_c%3D1.56%5C%20m.s%5E%7B-1%7D)
(b)![KE=5.3539\ J](https://tex.z-dn.net/?f=KE%3D5.3539%5C%20J)
(c-i)
in straight rightward direction.
(c-ii)![v_l=0\ m.s^{1}](https://tex.z-dn.net/?f=v_l%3D0%5C%20m.s%5E%7B1%7D)
(c-iii)
to the bottom of horizontal right.
(d-i)
to the horizontal right.
(d-ii)
horizontally left
(d-iii)
moving vertically downward
Explanation:
Given:
mass of hoop, ![m=2.2\ kg](https://tex.z-dn.net/?f=m%3D2.2%5C%20kg)
diameter of hoop, ![d=1.2\ m](https://tex.z-dn.net/?f=d%3D1.2%5C%20m)
angular speed of hoop, ![\omega=2.6\ rad.s^{-1}](https://tex.z-dn.net/?f=%5Comega%3D2.6%5C%20rad.s%5E%7B-1%7D)
So, time taken for 1 revolution(2π radians) of the hoop:
![T=\frac{2\pi}{2.6}\ s](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B2%5Cpi%7D%7B2.6%7D%5C%20s)
(a)
The center will move linearly in the right direction.
circumference of the hoop:
![c=\pi.d](https://tex.z-dn.net/?f=c%3D%5Cpi.d)
![c=\pi\times 1.2](https://tex.z-dn.net/?f=c%3D%5Cpi%5Ctimes%201.2)
<u>Now the speed of center:</u>
![v_c=\frac{c}{T}](https://tex.z-dn.net/?f=v_c%3D%5Cfrac%7Bc%7D%7BT%7D)
![v_c=\frac{\pi\times 1.2}{(\frac{2\pi}{2.6})}](https://tex.z-dn.net/?f=v_c%3D%5Cfrac%7B%5Cpi%5Ctimes%201.2%7D%7B%28%5Cfrac%7B2%5Cpi%7D%7B2.6%7D%29%7D)
![v_c=1.56\ m.s^{-1}](https://tex.z-dn.net/?f=v_c%3D1.56%5C%20m.s%5E%7B-1%7D)
(b)
Moment of inertia for ring about central axis:
![I=m.r^2](https://tex.z-dn.net/?f=I%3Dm.r%5E2)
where 'r' is the radius of the hoop.
![I=2.2\times 0.6^2](https://tex.z-dn.net/?f=I%3D2.2%5Ctimes%200.6%5E2)
![I=0.792\ kg.m^2](https://tex.z-dn.net/?f=I%3D0.792%5C%20kg.m%5E2)
∴Kinetic energy
![KE=\frac{1}{2} I.\omega^2+\frac{1}{2} m.v_c^2](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7D%20I.%5Comega%5E2%2B%5Cfrac%7B1%7D%7B2%7D%20m.v_c%5E2)
![KE=\frac{1}{2} 0.792\times 2.6^2+\frac{1}{2} 2.2\times 1.56^2](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7D%200.792%5Ctimes%202.6%5E2%2B%5Cfrac%7B1%7D%7B2%7D%202.2%5Ctimes%201.56%5E2)
![KE=5.3539\ J](https://tex.z-dn.net/?f=KE%3D5.3539%5C%20J)
(c) (i)
The highest point on the hoop will have the maximum velocity.
Given by:
![v_h=\omega \times d](https://tex.z-dn.net/?f=v_h%3D%5Comega%20%5Ctimes%20d)
![v_h=2.6\times 1.2](https://tex.z-dn.net/?f=v_h%3D2.6%5Ctimes%201.2)
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}](https://tex.z-dn.net/?f=v_l%3D0%5C%20m.s%5E%7B1%7D)
(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}](https://tex.z-dn.net/?f=v_r%3D%5Csqrt%7Bv_c%5E2%2Bv_d%5E2%7D)
here:
is the downward component.
![v_d=r.\omega](https://tex.z-dn.net/?f=v_d%3Dr.%5Comega)
![v_d=1.56\ m.s^{-1}](https://tex.z-dn.net/?f=v_d%3D1.56%5C%20m.s%5E%7B-1%7D)
![\therefore v_r=\sqrt{1.56^2+1.56^2}](https://tex.z-dn.net/?f=%5Ctherefore%20v_r%3D%5Csqrt%7B1.56%5E2%2B1.56%5E2%7D)
![v_r=2.2062\ m.s^{-1}](https://tex.z-dn.net/?f=v_r%3D2.2062%5C%20m.s%5E%7B-1%7D)
![tan\theta=\frac{1.56}{1.56}](https://tex.z-dn.net/?f=tan%5Ctheta%3D%5Cfrac%7B1.56%7D%7B1.56%7D)
to the bottom of horizontal right.
When the observer is moving in the same direction with
velocity:
(d) i
Then,
![v_h=\omega \times r](https://tex.z-dn.net/?f=v_h%3D%5Comega%20%5Ctimes%20r)
![v_h=2.6\times 0.6](https://tex.z-dn.net/?f=v_h%3D2.6%5Ctimes%200.6)
to the horizontal right.
(d) ii
The bottom point of hoop will seem to move horizontally left with velocity:
![v_l=r.\omega](https://tex.z-dn.net/?f=v_l%3Dr.%5Comega)
![v_l=0.6\times 2.6](https://tex.z-dn.net/?f=v_l%3D0.6%5Ctimes%202.6)
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](https://tex.z-dn.net/?f=v_r%3Dr.%5Comega)
![v_r=0.6\times 2.6](https://tex.z-dn.net/?f=v_r%3D0.6%5Ctimes%202.6)
![v_r=1.56\ m.s^{-1}](https://tex.z-dn.net/?f=v_r%3D1.56%5C%20m.s%5E%7B-1%7D)