Respuesta :
Answer:
(a)[tex]v_c=1.56\ m.s^{-1}[/tex]
(b)[tex]KE=5.3539\ J[/tex]
(c-i)[tex]v_h=3.12\ m.s^{-1}[/tex] in straight rightward direction.
(c-ii)[tex]v_l=0\ m.s^{1}[/tex]
(c-iii)[tex]v_r=2.2062\ m.s^{-1}[/tex] [tex]\theta=45^{\circ}[/tex] to the bottom of horizontal right.
(d-i) [tex]v_h=1.56\ m.s^{-1}[/tex] to the horizontal right.
(d-ii)[tex]v_l=1.56\ m.s^{-1}[/tex] horizontally left
(d-iii)[tex]v_r=1.56\ m.s^{-1}[/tex] moving vertically downward
Explanation:
Given:
mass of hoop, [tex]m=2.2\ kg[/tex]
diameter of hoop, [tex]d=1.2\ m[/tex]
angular speed of hoop, [tex]\omega=2.6\ rad.s^{-1}[/tex]
So, time taken for 1 revolution(2π radians) of the hoop:
[tex]T=\frac{2\pi}{2.6}\ s[/tex]
(a)
The center will move linearly in the right direction.
circumference of the hoop:
[tex]c=\pi.d[/tex]
[tex]c=\pi\times 1.2[/tex]
Now the speed of center:
[tex]v_c=\frac{c}{T}[/tex]
[tex]v_c=\frac{\pi\times 1.2}{(\frac{2\pi}{2.6})}[/tex]
[tex]v_c=1.56\ m.s^{-1}[/tex]
(b)
Moment of inertia for ring about central axis:
[tex]I=m.r^2[/tex]
where 'r' is the radius of the hoop.
[tex]I=2.2\times 0.6^2[/tex]
[tex]I=0.792\ kg.m^2[/tex]
∴Kinetic energy
[tex]KE=\frac{1}{2} I.\omega^2+\frac{1}{2} m.v_c^2[/tex]
[tex]KE=\frac{1}{2} 0.792\times 2.6^2+\frac{1}{2} 2.2\times 1.56^2[/tex]
[tex]KE=5.3539\ J[/tex]
(c) (i)
The highest point on the hoop will have the maximum velocity.
Given by:
[tex]v_h=\omega \times d[/tex]
[tex]v_h=2.6\times 1.2[/tex]
[tex]v_h=3.12\ m.s^{-1}[/tex] in straight rightward direction.
(c) (ii)
Lowest point n the hoop will seem stationary for an observer on the ground.
[tex]v_l=0\ m.s^{1}[/tex]
(c) (iii)
Velocity of the right-most point of the loop.
This velocity will have 2 components horizontal right and vertical down.
[tex]v_r=\sqrt{v_c^2+v_d^2}[/tex]
here: [tex]v_d[/tex] is the downward component.
[tex]v_d=r.\omega[/tex]
[tex]v_d=1.56\ m.s^{-1}[/tex]
[tex]\therefore v_r=\sqrt{1.56^2+1.56^2}[/tex]
[tex]v_r=2.2062\ m.s^{-1}[/tex]
[tex]tan\theta=\frac{1.56}{1.56}[/tex]
[tex]\theta=45^{\circ}[/tex] to the bottom of horizontal right.
When the observer is moving in the same direction with [tex]v_c[/tex] velocity:
(d) i
Then,
[tex]v_h=\omega \times r[/tex]
[tex]v_h=2.6\times 0.6[/tex]
[tex]v_h=1.56\ m.s^{-1}[/tex] to the horizontal right.
(d) ii
The bottom point of hoop will seem to move horizontally left with velocity:
[tex]v_l=r.\omega[/tex]
[tex]v_l=0.6\times 2.6[/tex]
[tex]v_l=1.56\ m.s^{-1}[/tex] 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:
[tex]v_r=r.\omega[/tex]
[tex]v_r=0.6\times 2.6[/tex]
[tex]v_r=1.56\ m.s^{-1}[/tex]
A rotating hoop in translational motion, without slipping, moves fastest at
the top.
The correct responses are;
(a) 1.56 m/s
(b) 5.35392 J
(c) (i) The velocity at the highest point on the hoop is 3.12·i
(ii) The velocity vector at the lowest point on the hoop is 0
(iii) The velocity vector midway on the right is 1.56·i - 1.56·j
(d) (i) The (relative) velocity vector at the highest point is 1.56·i
(ii) The (relative) vector velocity at the lowest point is -1.56·i
(iii) The velocity vector at the mid right hoop is -1.56·j
Reasons:
Given information;
Mass of the hoop = 2.20 kg
The hoop diameter, D = 1.20 m
Hoop rotational speed, ω = 2.60 rad/s
Required:
(a) The speed of the center of the hoop
Solution:
The center of the hoop has only translational motion, v, given as follows;
[tex]v = \omega \times Radius, \ r[/tex]
[tex]r=\dfrac{D}{2} = \dfrac{1.2}{2} = 0.6[/tex]
The radius of the hoop, r = 0.6 m
Therefore;
[tex]v_{htr}= \omega \times \dfrac{D}{2}[/tex]
Therefore;
[tex]v_{htr}= 2.60 \times \dfrac{1.2}{2} = 1.56[/tex]
The speed of the center of the hoop, [tex]v_{htr}[/tex] = 1.56 m/s
(b) The total kinetic energy of the hoop, K.E., is given as follows;
[tex]K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v_{htr}^2 + \dfrac{1}{2} \cdot I \cdot \omega^2}[/tex]
Therefore
[tex]I = \ m\cdot r^2[/tex]
[tex]K.E. = \dfrac{1}{2} \times 2.20 \cdot 1.56^2 + \dfrac{1}{2} \times 2.2 \times 0.6^2 \times2.6^2 = 5.35392[/tex]
The total kinetic energy of the hoop, K.E. = 5.35392 J
(c) The velocity vectors as viewed by a someone at rest on the ground are;
(i) The tangential velocity vector of the top of the hoop is tangential to the
top, therefore, it is in the direction of the hoop.
The magnitude of the velocity, |[tex]\mathbf{v_{tan,t}}[/tex]| = ω×r
∴ |[tex]\mathbf{v_{tan, t}}[/tex]| = 2.60 × 0.6 = 1.56
Direction = Positive x-direction (Direction of motion of the hoop)
The velocity vector, [tex]\mathbf{v_{tan, t}}[/tex] = 1.56·i
The velocity due to the translational of the hoop, [tex]\mathbf{v_{htr}}[/tex] = 1.56·i
The velocity at the top of the hoop, [tex]\mathbf{v_{t}}[/tex] = [tex]\mathbf{v_{tan, t}}[/tex] + [tex]\mathbf{v_{htr}}[/tex]
[tex]\mathbf{v_{tan, t}}[/tex] = [tex]\mathbf{v_{htr}}[/tex]
∴ [tex]\mathbf{v_{t}}[/tex] = 2·[tex]\mathbf{v_{htr}}[/tex]
∴ [tex]\mathbf{v_{t}}[/tex] = 1.56·i + 1.56·i = 3.12·i
[tex]\mathbf{v_{t}}[/tex] = 3.12·i
The magnitude of the velocity is 3.12 m/s, the direction is forward (θ = 0°)
(ii) The direction of the velocity at the bottom
[tex]\mathbf{v_{tan, b}}[/tex] = [tex]-\mathbf{v_{tan, t}}[/tex]
Therefore;
The magnitude of the velocity at the bottom of the hoop, [tex]\mathbf{v_{b}}[/tex]
[tex]\mathbf{v_{b}}[/tex] = [tex]\mathbf{v_{tan, b}}[/tex] + [tex]\mathbf{v_{htr}}[/tex] = [tex]-\mathbf{v_{tan, t}}[/tex] + [tex]\mathbf{v_{htr}}[/tex]
[tex]\mathbf{v_{b}}[/tex] = -1.56·i + 1.56·i = 0
The velocity vector at the lowest point on the hoop, [tex]\mathbf{v_{b}}[/tex] = 0
The magnitude of the velocity vector at the bottom, [tex]\mathbf{v_{b}}[/tex] = 0
(iii) The velocity midway between the top and bottom of the hoop on the
right is given as follows;
The tangential velocity, [tex]\mathbf{v_{tan, m}}[/tex] = 1.56·j (acting downwards)
Sum of the velocity, [tex]\mathbf{v_{m}}[/tex] = [tex]\mathbf{v_{tan, m}}[/tex] + [tex]\mathbf{v_{htr}}[/tex]
∴ [tex]\mathbf{v_{m}}[/tex] = -1.56·j + 1.56·i
The velocity vector midway, [tex]\mathbf{v_{m}}[/tex] = 1.56·i - 1.56·j
The magnitude = √(1.56² + 1.56²) = (√2)·1.56 m/s ≈ 2.206 m/s
The direction of the velocity, θ = [tex]arctan \left(-\dfrac{1.56}{1.56} \right)[/tex] = -45°
The direction of the velocity, θ = -45°
(d) The velocity at the above points relative to someone moving along with
the hoop, with a speed, v = [tex]\mathbf{v_{htr}}[/tex] are;
Relative translational velocity, [tex]\mathbf{v_{rhtr}}[/tex] = v - [tex]\mathbf{v_{htr}}[/tex] = 0
(i) The velocity vector at the highest point [tex]\mathbf{v_{t}}[/tex] = [tex]\mathbf{v_{tan, t}}[/tex] + [tex]\mathbf{v_{rhtr}}[/tex] = [tex]\mathbf{v_{tan, t}}[/tex] = 1.56·i
The velocity vector at the highest point [tex]\mathbf{v_{t}}[/tex] = 1.56·i
The magnitude = 1.56 m/s
The direction = 0°
(ii) The velocity at the lowest point [tex]\mathbf{v_{b}}[/tex] = [tex]\mathbf{v_{tan, b}}[/tex] + [tex]\mathbf{v_{rhtr}}[/tex] = [tex]-\mathbf{v_{tan, t}}[/tex] + 0 = -1.56·i
The velocity vector at the lowest point [tex]\mathbf{v_{b}}[/tex] = -1.56·i (acting backwards)
The magnitude = 1.56 m/s
The direction = 180°
(iii) Velocity at the mid right hoop, [tex]\mathbf{v_{m}}[/tex] = [tex]\mathbf{v_{tan, m}}[/tex] + [tex]\mathbf{v_{rhtr}}[/tex] = -1.56·j
Velocity vector at the mid right hoop, [tex]\mathbf{v_{m}}[/tex] = -1.56·j (acting downwards)
The magnitude = 1.56 m/s
The direction = -90°
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