Answer:
radial acceleration is 41.8 m / s²
Explanation:
The acceleration for circular motion is
a = v² / r
They also give us the X and Y position where the body falls when the rope breaks, let's write the projectile launch equations
x = vox t
y = v₀ₓ t - ½ g t2
Since the circle is horizontally the v₀ₓ is zero (v₀ₓ = 0)
x = v₀ₓ t
t = x / v₀ₓ
y = - ½ g t²
Let's replace and calculate the initial velocity on the X axis
y = - ½ g (x / vox)²
v₀ₓ = √ (g x² / 2 y)
v₀ₓ = √ [- (-9.8) 1.6² / (2 1.00)]
v₀ₓ = 3.54 m / s
This is the horizontal velocity, but since it circle is in horizontal position it is also the velocity of the body at the point of rupture.
Now we can calculate the radial acceleration
a = v² / r
a = 3.54² / 0.300
a = 41.8 m / s²
The formula relating acceleration and angular velocity is:
a = ω^2 r
where a is acceleration, ω is angular velocity and r is
radius
But the angular velocity ω is constant all throughout the
disk therefore:
a1 / r1 = a2 / r2
So at points:
<span>r1 = 0.0130 m ->
a1 = 393 m/s^2</span>
<span>r2 = 0.0884 m ->
a2 = ?</span>
393 / 0.0130 = a2 / 0.0884
<span>a2 = 2,672.4 m/s^2</span>
Force [kgms^-2] = mass [kg] x acceleration [ms^-2]
Work = force x distance
Work = [kgms^-2] x [m]
Work = [kgm^2s^-2]
Answer:
The new height the ball will reach = (1/4) of the initial height it reached.
Explanation:
The energy stored in any spring material is given as (1/2)kx²
This energy is converted to potential energy, mgH, of the ball at its maximum height.
If the initial height reached is H
And the initial compression of the spring = x
So, mgH = (1/2)kx²
H = kx²/2mg
The new compression, x₁ = x/2
New energy of loaded spring = (1/2)kx₁²
And the new potential energy = mgH₁
mgH₁ = (1/2)kx₁²
But x₁ = x/2
mgH₁ = (1/2)k(x/2)² = kx²/8
H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)
