A solid sphere of radius 23 cm is positioned at the top of an incline that makes 23 degree angle with the horizontal. This initial position of the sphere is a vertical distance 3 m above its position when at the bottom of the incline. The sphere is released and moves down the incline.
Find the speed of the sphere when it reaches the bottom of the incline if it rolls without slipping. The acceleration of gravity is 9.8 m/s^2 and the moment of inertia of a sphere with respect to an axis through its center is 2/5 MR^2.
Find the speed of the sphere if it reaches the bottom of the incline by slipping frictionlessly without rolling.
To find the speed of the sphere when it reaches the bottom of the incline, we'll analyze two scenarios: rolling without slipping and slipping frictionlessly without rolling.
1. Rolling without slipping:
In this scenario, the sphere will roll down the incline without slipping, which means that its linear and angular motion are connected. We can use the concept of conservation of energy to find the speed at the bottom of the incline.
First, let's calculate the change in gravitational potential energy of the sphere. The initial position of the sphere is 3 m above its position at the bottom of the incline, so the change in height is Δh = 3 m.
The change in gravitational potential energy is given by ΔPE = m * g * Δh, where m is the mass of the sphere and g is the acceleration due to gravity.
Next, let's calculate the change in kinetic energy of the sphere. Since it is rolling without slipping, the total kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy.
The translational kinetic energy (KEt) is given by KEt = (1/2) * m * v^2, where v is the linear velocity of the sphere.
The rotational kinetic energy (KEr) is given by KEr = (1/2) * I * ω^2, where I is the moment of inertia of the sphere and ω is the angular velocity.
For a solid sphere, the moment of inertia (I) about an axis through its center is (2/5) * m * r^2, where r is the radius of the sphere.
Since the sphere is rolling without slipping, the linear velocity (v) is related to the angular velocity (ω) through the equation v = ω * r.
Now, let's equate the change in gravitational potential energy to the change in kinetic energy:
m * g * Δh = (1/2) * m * v^2 + (1/2) * I * ω^2
Simplify the expression using the moment of inertia formula and the relation between ω and v:
m * g * Δh = (1/2) * m * v^2 + (1/2) * ((2/5) * m * r^2) * ((v / r)^2)
Cancel out the common term of "m":
g * Δh = (1/2) * v^2 + (1/5) * v^2
Simplify further:
g * Δh = (7/10) * v^2
Rearrange the equation to solve for v:
v^2 = (10/7) * g * Δh
v = sqrt((10/7) * g * Δh)
Substitute the given values:
v = sqrt((10/7) * 9.8 m/s^2 * 3 m)
Solve for v using a calculator:
v ≈ 7.48 m/s
Therefore, the speed of the sphere when it reaches the bottom of the incline, rolling without slipping, is approximately 7.48 m/s.
2. Slipping frictionlessly without rolling:
In this scenario, the sphere will slide down the incline without rolling or experiencing any rotational motion.
To find the speed at the bottom of the incline, we can use the concept of conservation of energy again.
The change in gravitational potential energy is the same as before, given by ΔPE = m * g * Δh.
However, since the sphere is not rolling, there is no contribution from rotational kinetic energy. Hence, the total kinetic energy is only due to translational motion.
The translational kinetic energy is given by KEt = (1/2) * m * v^2.
Let's equate the change in gravitational potential energy to the translational kinetic energy:
m * g * Δh = (1/2) * m * v^2
Cancel out the common term of "m":
g * Δh = (1/2) * v^2
Rearrange the equation to solve for v:
v^2 = 2 * g * Δh
v = sqrt(2 * g * Δh)
Substitute the given values:
v = sqrt(2 * 9.8 m/s^2 * 3 m)
Solve for v using a calculator:
v ≈ 7.67 m/s
Therefore, the speed of the sphere when it reaches the bottom of the incline, slipping frictionlessly without rolling, is approximately 7.67 m/s.