A 290-N sphere 0.20 m in radius rolls without slipping 6.0 m down a ramp that is inclined at 31° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?
Use conservation of energy and include the rotational kinetic energy, (1/2) I w^2. They should have told you whether the sphere has uniform density or is hollow inside. It makes a difference.
For a uniform-density sphere,
M g H = (1/2) M V^2 + (1/2) I w^2
= (1/2) M V^2 + (1/2)(2/5)M R^2*(V/R)^2
= (7/10)M V^2
M cancels out. H is the elevation loss, 6.0 sin 31 = 3.09 meters
Solve for V, the final velocity. For the final angular speed, use
w = V/R
rdgh
To find the angular speed of the sphere at the bottom of the slope, we can use the principles of conservation of energy and rotational motion.
Step 1: Find the gravitational potential energy at the top of the slope.
The potential energy (PE) can be calculated using the formula: PE = m * g * h
where m is the mass of the sphere and g is the acceleration due to gravity.
Given that the weight (W) is equal to mass (m) multiplied by the gravitational acceleration (g), and the weight is given as 290 N, we can find the mass of the sphere: m = W / g.
The gravitational acceleration (g) is approximately 9.8 m/s^2.
The height difference (h) is equal to the distance down the slope since it starts from rest.
h = 6.0 m.
Step 2: Find the final kinetic energy at the bottom of the slope.
The kinetic energy (KE) is given by the formula: KE = (1/2) * I * ω^2
where I is the moment of inertia of the sphere and ω is the angular velocity.
The moment of inertia of a solid sphere is (2/5) * m * r^2, where r is the radius.
Given that the radius (r) is 0.20 m, we can find the moment of inertia (I) using the mass (m) we calculated in Step 1.
Step 3: Equate the potential energy at the top of the slope to the final kinetic energy at the bottom of the slope.
PE = KE
Step 4: Solve for the angular velocity (ω).
We now have the equation: m * g * h = (1/2) * ((2/5) * m * r^2) * ω^2
Simplifying the equation, we get: g * h = (1/5) * r^2 * ω^2
Solving for ω, we have: ω = sqrt((5 * g * h) / r^2)
Let's substitute the given values into the formula to find the angular speed.
g = 9.8 m/s^2 (gravitational acceleration)
h = 6.0 m (height difference)
r = 0.20 m (radius of the sphere)
ω = sqrt((5 * 9.8 m/s^2 * 6.0 m) / (0.20 m)^2)
Calculating this, we get:
ω = sqrt((5 * 9.8 * 6.0) / (0.04))
ω = sqrt(294 / 0.04)
ω = sqrt(7350)
ω ≈ 85.74 rad/s (rounded to two decimal places)
Therefore, the angular speed of the sphere at the bottom of the slope is approximately 85.74 rad/s.
To find the angular speed of the sphere at the bottom of the slope, we need to consider the conservation of mechanical energy.
The mechanical energy of the system is conserved when there are no external forces doing work on it. In this case, we assume no energy losses due to friction or air resistance.
The mechanical energy of the system is the sum of kinetic energy (KE) and potential energy (PE). Initially, the sphere is at rest, so it has no kinetic energy, and all the energy is in the form of potential energy due to its position on the ramp. At the bottom of the slope, all the potential energy is converted to kinetic energy, and there is no change in height, so there is no potential energy.
The potential energy of the sphere at the starting position can be calculated using the formula:
PE = m * g * h
where m is the mass of the sphere, g is the acceleration due to gravity, and h is the vertical height of the starting position.
Since the sphere is on an inclined plane, the vertical height can be calculated as:
h = sin(θ) * L
where θ is the angle of inclination of the ramp and L is the length of the ramp.
Given:
m = 290 N (convert to kg by dividing by g ≈ 9.8 m/s²)
r = 0.20 m (radius of the sphere)
θ = 31° (angle of the ramp)
L = 6.0 m (length of the ramp)
First, we need to calculate the mass of the sphere:
m = F / g
= 290 N / 9.8 m/s²
≈ 29.59 kg
Next, calculate the vertical height:
h = sin(31°) * 6.0 m
≈ 3.08 m
Then, calculate the potential energy at the starting position:
PE = m * g * h
≈ 29.59 kg * 9.8 m/s² * 3.08 m
≈ 897 J
At the bottom of the slope, all the potential energy is converted into kinetic energy, and there is no change in height:
KE = PE = 897 J
The kinetic energy of a rolling sphere is given by:
KE = (1/2) * I * ω²
where I is the moment of inertia and ω is the angular speed.
For a solid sphere rolling without slipping, the moment of inertia is given by:
I = (2/5) * m * r²
Substituting the values:
897 J = (1/2) * (2/5) * 29.59 kg * (0.20 m)² * ω²
Simplifying:
897 J = (1/5) * 2.95 kg * (0.20 m)² * ω²
897 J = 0.118 J * ω²
Dividing both sides by 0.118 J:
ω² = 897 J / 0.118 J
ω² ≈ 7605
Taking the square root of both sides:
ω ≈ √7605
ω ≈ 87.2 rad/s
So, the angular speed of the sphere at the bottom of the slope is approximately 87.2 rad/s.