A solid sphere of radius r is gently placed on rough horizontal ground with an initial angular speed and no linear velocity if the coefficient of friction is find the time when the slipping stops in addition state the linear velocity and angular velocity at the end of slipping
To find the time when the slipping stops, as well as the linear and angular velocities at that point, we need to consider the forces acting on the sphere.
1. The only horizontal force acting on the sphere is the frictional force (f). This force opposes the motion and causes the slipping. The frictional force can be calculated using the equation f = μ * N, where μ is the coefficient of friction and N is the normal force.
2. The normal force (N) is equal to the weight of the sphere, which can be calculated as N = m * g, where m is the mass of the sphere and g is the acceleration due to gravity.
3. The torque produced by the frictional force causes a deceleration in the rotational motion of the sphere. The torque (τ) can be calculated as τ = f * r, where r is the radius of the sphere.
4. The net torque (τ_net) is equal to the moment of inertia (I) times the angular acceleration (α). In this case, the net torque is equal to the torque produced by the frictional force. Thus, τ_net = τ.
5. The moment of inertia of a solid sphere about its center is given by I = (2/5) * m * r^2, where m is the mass of the sphere, and r is the radius of the sphere.
6. The angular acceleration (α) can be calculated as α = τ_net / I.
7. When the slipping stops, the linear velocity (v) is equal to the product of the radius (r) and angular velocity (ω). Therefore, v = r * ω.
To find the time when the slipping stops, we can use the equation of motion: v = u + a * t, where u is the initial velocity, a is the acceleration, and t is the time.
Let's summarize the steps to solve the problem:
Step 1: Calculate the normal force N = m * g.
Step 2: Calculate the frictional force f = μ * N.
Step 3: Calculate the torque τ = f * r.
Step 4: Set the torque equal to the moment of inertia times the angular acceleration: τ_net = τ.
Step 5: Calculate the moment of inertia I = (2/5) * m * r^2.
Step 6: Calculate the angular acceleration α = τ_net / I.
Step 7: Calculate the linear velocity at the end of slipping v = r * ω.
Step 8: Set the linear velocity equal to zero and solve for the time t when slipping stops: 0 = r * ω + a * t.
To find the time when the slipping stops, as well as the linear velocity and angular velocity at the end of slipping, we can use the equations of rotational motion and friction.
First, let's analyze the forces acting on the sphere during the slipping phase:
1. Weight (mg): This force always acts downward and is equal to the mass (m) of the sphere multiplied by the acceleration due to gravity (g).
2. Normal force (N): This force acts perpendicular to the ground and is equal to the weight of the sphere, so N = mg.
3. Frictional force (f): This force opposes the motion of the sphere and acts parallel to the ground. It is given by f = μN, where μ is the coefficient of friction.
During the slipping phase, the frictional force provides the torque necessary to decelerate the sphere's rotation until it eventually stops slipping. The torque (τ) is defined as τ = Iα, where I is the moment of inertia of the sphere and α is the angular acceleration.
The moment of inertia of a solid sphere is given by I = (2/5)mr^2, where m is the mass of the sphere and r is the radius.
Since the angular acceleration is constant during slipping, we can use the kinematic equation for rotational motion: ω^2 = ω_0^2 + 2αθ, where ω is the final angular velocity, ω_0 is the initial angular velocity, α is the angular acceleration, and θ is the angular displacement.
Since the sphere starts from rest and decelerates until slipping stops, the final angular velocity ω will be zero. Therefore, ω^2 = 0, and the equation becomes ω_0^2 = -2αθ.
The angular displacement θ can be calculated using the equation θ = ω_0t + (1/2)αt^2, where t is the time it takes for the slipping to stop.
Let's solve for t:
0 = ω_0^2 + 2αθ
0 = ω_0^2 + 2α(ω_0t + (1/2)αt^2)
Substituting the expression 2αθ from the first equation into the second equation gives:
0 = ω_0^2 + 2α(ω_0t + (1/2)αt^2)
0 = ω_0^2 + 2ω_0tα + α^2t^2
This equation is quadratic in t, so we can solve it using the quadratic formula:
t = (-2ω_0α ± √(4ω_0^2α^2 - 4α^2(ω_0^2))) / (2α^2)
t = (-2ω_0α ± √(4ω_0^2α^2 - 4α^2ω_0^2)) / (2α^2)
t = (-2ω_0α ± √(4α^2(ω_0^2 - α^2))) / (2α^2)
t = (-2ω_0α ± 2α√(ω_0^2 - α^2)) / (2α^2)
t = (-ω_0 ± √(ω_0^2 - α^2)) / α
Since t represents time, we can ignore the negative solution:
t = (-ω_0 + √(ω_0^2 - α^2)) / α
To find the linear velocity (v) at the end of slipping, we can use the equation v = ωr, where r is the radius of the sphere.
v = ωr = (αt)r = (α[(-ω_0 + √(ω_0^2 - α^2)) / α])r
v = (-ω_0 + √(ω_0^2 - α^2)) r
Finally, the angular velocity (ω) at the end of slipping is zero, as the sphere stops rotating.
Therefore, at the end of slipping:
- The time when the slipping stops is t = (-ω_0 + √(ω_0^2 - α^2)) / α.
- The linear velocity is v = (-ω_0 + √(ω_0^2 - α^2)) r.
- The angular velocity is ω = 0.
Note: To find the values for α and ω_0, additional information or equations might be needed, such as the force applied to start the sphere's rotation or the moment of inertia of the sphere.