A solid sphere of radius 3 cm and mass of 100 g rolls down a 30 degree incline from a height of 1 m. A hoop of the same mass and radius rolls down the same incline from the same height. What is the linear velocity of each body at the bottom of the incline? From what height would the body with the larger moment of inertia have to roll down from so that the two linear velocities at the bottom of the incline would be the same?
Do not forget to include rotational kinetic energy when you apply the law of conservation of energy.
It is very unlikely that a hoop "of the same radius" will have the same mass as the sphere. It also turns out that the mass and radius do not affect the result. The sphere wil travel faster because the coefficient in front of MR^2 in the moment of inertia term is less. The M and R terms cancel out when calculating the speed.
This is another poorly formulated question. You are wasting your time taking this course from such an incompetent instructor or school
To find the linear velocity of each body at the bottom of the incline, we can use the principle of conservation of energy. The initial potential energy is converted into the kinetic energy of the rolling bodies.
For the solid sphere:
1. The potential energy at the top of the incline can be calculated as mgh, where m is the mass (100 g), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (1 m).
Potential energy = (0.1 kg) x (9.8 m/s^2) x (1 m) = 0.98 J
2. The final kinetic energy of a rolling body is given by the equation (1/2)mv^2 + (1/2)Iω^2, where v is the linear velocity, I is the moment of inertia, and ω is the angular velocity. For a solid sphere rolling without slipping, the relationship between linear velocity and angular velocity is v = ωr, where r is the radius of the sphere.
The moment of inertia for a solid sphere is (2/5)mr^2, where m is the mass and r is the radius. For our sphere, I = (2/5)(0.1 kg)(0.03 m)^2 = 0.00036 kg·m^2
Substituting the values into the equation, we get 0.98 J = (1/2)(0.1 kg)v^2 + (1/2)(0.00036 kg·m^2)(v/0.03 m)^2
Simplifying, we have 0.98 J = (0.05 kg)v^2 + 0.12v^2
Combining like terms, we get 0.98 J = 0.17v^2
3. Solving for v, we find v^2 = (0.98 J) / (0.17 kg) which gives us v ≈ 2.86 m/s (rounded to two decimal places)
So, the linear velocity of the solid sphere at the bottom of the incline is approximately 2.86 m/s.
Now let's calculate the linear velocity for the hoop:
Since a hollow hoop has all the mass concentrated at its circumference, its moment of inertia is given by I = mr^2, where m is the mass (0.1 kg) and r is the radius (0.03 m). Therefore, the moment of inertia for the hoop is I = (0.1 kg)(0.03 m)^2 = 0.00009 kg·m^2.
Using the same conservation of energy principle, we can set up the equation for the hoop:
0.98 J = (1/2)(0.1 kg)v_hoop^2 + (1/2)(0.00009 kg·m^2)(v_hoop/0.03 m)^2
Simplifying, we get 0.98 J = 0.05v_hoop^2 + 0.09v_hoop^2
Combining like terms, we have 0.98 J = 0.14v_hoop^2
Solving for v_hoop, we find v_hoop^2 = (0.98 J) / (0.14 kg) which gives us v_hoop ≈ 3.52 m/s (rounded to two decimal places)
Therefore, the linear velocity of the hoop at the bottom of the incline is approximately 3.52 m/s.
To find the height from which the body with the larger moment of inertia would have to roll down so that the two linear velocities at the bottom of the incline are the same, we can equate the kinetic energy equations for the two bodies.
For the solid sphere:
(1/2)(0.1 kg) v_sphere^2 + (1/2)(0.00036 kg·m^2)(v_sphere/0.03 m)^2 = K (where K is the kinetic energy)
For the hoop:
(1/2)(0.1 kg) v_hoop^2 + (1/2)(0.00009 kg·m^2)(v_hoop/0.03 m)^2 = K
Since we want the two linear velocities to be the same, we can set v_sphere = v_hoop = v
By equating the two kinetic energy equations, we have:
(1/2)(0.1 kg) v^2 + (1/2)(0.00036 kg·m^2)(v/0.03 m)^2 = (1/2)(0.1 kg) v^2 + (1/2)(0.00009 kg·m^2)(v/0.03 m)^2
Simplifying, we get:
(1/2)(0.00036 kg·m^2)(v/0.03 m)^2 = (1/2)(0.00009 kg·m^2)(v/0.03 m)^2
Dividing both sides by (0.5)(0.00009 kg·m^2), we have:
(v/0.03 m)^2 = (v/0.03 m)^2
This equation shows that the height does not affect the outcome; thus, the body with the larger moment of inertia would need to roll down from the same height of 1 m in order for the two linear velocities at the bottom of the incline to be the same.