Answer:
Explanation:
The problem is related to rotational motion . So we shall find out rotational kinetic energy .
K E = 1/2 x I ω²
ω is the final angular velocity
Moment of inertial of the disk
I ₁ = 1/2 m r²
= .5 x 165 x 2.93²
= 708.25 kgm²
Moment of inertial of the person
I₂ = mr²
= 62.5 x 2.93²
= 536.55 kgm²
ω₂ = v / R
= 3.11 / 2.93 rad /s
At the time of jumping , law of conservation of angular momentum will apply
I₁ ω₁ + I₂ω₂ = (I₁ + I₂)ω
708.25 x0.691 + 536.55 x ( 3.11 / 2.93 ) = ( 708.25 + 536.55 ) ω
ω = 0 .85 rad/ s
K E = 1/2 x I ω²
= .5 x ( 708.25 + 536.55 ) ( .85 )²
449.68 J
Answer:
<em>(a) t = 4.52 sec</em>
<em>(b) X = 1,156.49 m</em>
Explanation:
<u>Horizontal Launching
</u>
If an object is launched horizontally, its initial speed is zero in the y-coordinate and the horizontal component of the velocity 
remains the same in time. The distance x is computed as
.
(a)
The vertical component of the velocity 
starts from zero and gradually starts to increase due to the acceleration of gravity as follows
This means the vertical height is computed by
Where 
is the initial height. Our fighter bomber is 100 m high, so we can compute the time the bomb needs to reach the ground by solving the above equation for t knowing h=0
(b)
We now compute the horizontal distance knowing 
Answer: the same direction I.e to the left.
Explanation:
The component perpendicular to the contact surface is such that will stop the relative motion and, in case of elastic collision like here, return the system to the same kinetic energy. So ball hitting immovable surface will have the same speed (magnitude of velocity) as before the collision.
There will also be parallel force caused by friction, but it has to be treated separately for two reasons:
The perpendicular force is limited to coefficient of friction times the normal force. If that is not enough to stop the ball, it will skid on the surface.The perpendicular force, and this depends on the specific geometry, does not pass through the centre of mass of the ball. Therefore it imparts a moment on the ball that causes it to start rotating. And once the ball is rotating so that the point of contact is stationary, there is no momentum to cause any friction force anymore and the friction force disappears and stops decelerating the ball.
So what happens is that the vertical component of the velocity will be reversed, while the horizontal component will be somewhat reduced with the corresponding amount of kinetic energy transferred to energy of rotation. The rotation will always eliminate the friction force before the horizontal component of velocity is zeroed, so the ball will always continue in the same direction, just a bit slower.
If you instead threw an elastic box (which could not start rotating freely) it could actually bounce back.
Answer:
θ = 12.95º
Explanation:
For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height
Let's start by finding the speed of the bar plus clay ball system, using amount of momentum
The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)
Initial before the crash
p₀ = m v₀
Final after the crash before starting the movement
= (m + M) v
p₀ =
m v₀ = (m + M) v
v = v₀ m / (m + M)
v = 2.0 0.015 / (0.015 +0.080)
v = 0.316 m / s
With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy
Lower
Em₀ = K = ½ (m + M) v²
Higher
= U = (m + M) g y
Em₀ =
½ (m + M) v² = (m + M) g y
y = ½ v² / g
y = ½ 0.316² / 9.8
y = 0.00509 m
Let's look for the angle the height from the pivot point is
L = 0.40 / 2 = 0.20 cm
The distance that went up is
y = L - L cos θ
cos θ = (L-y) / L
θ = cos⁻¹ (L-y) / L
θ = cos⁻¹-1 ((0.20 - 0.00509) /0.20)
θ = 12.95º
