A uniform solid sphere rolls along a horizontal frictionless surface at 35 m/s and makes a smooth transition onto a frictionless incline having an angle of 300. How far up the incline does the sphere roll when it comes to a momentary stop? Note for a sphere I= 2/5Mr2, where M is the mass of the sphere and r is the radius of the sphere.
Its an angle of 30 not 300
Original ke = (1/2)I omega^2 + (1/2)m v^2
omega = v/r = 35/r
ke=(1/2)(2/5)mr^2(35^2)/r^2+(1/2)m (35)^2
= (1/2)m [(2/5)+1](35)^2
= (1/2) m (1.4)(35)^2
distance up = h = L sin 20 = L/2
potential energy stopped at top = original ke
m g h = m g L/2 = (1/2)m(1.4)(35)^2
note m cancels
g = 9.81 on earth
To find how far up the incline the sphere rolls when it comes to a momentary stop, we can use the conservation of mechanical energy and the rotational motion equations.
In this problem, the initial kinetic energy of the sphere is converted into potential energy as it moves up the incline. At the moment it comes to a stop, all of its initial kinetic energy is transformed into potential energy.
Let's calculate the initial kinetic energy of the sphere:
Kinetic energy (KE) = 0.5 * mass * velocity^2
Given that the velocity of the sphere is 35 m/s, we need to know the mass and radius of the sphere. Could you please provide those values?
To find the distance the sphere rolls up the incline before coming to a momentary stop, we can use the principle of conservation of mechanical energy.
The mechanical energy of the sphere is conserved because there is no friction or external force acting on it. The initial mechanical energy of the sphere is equal to its final mechanical energy when it comes to a stop.
The initial mechanical energy is the sum of the kinetic energy and potential energy at the bottom of the incline:
Initial Mechanical Energy = Kinetic Energy + Potential Energy
The kinetic energy of the rolling sphere is given by the formula:
Kinetic Energy = 1/2 * I * ω^2
Where I is the moment of inertia and ω is the angular velocity of the sphere. For a rolling sphere, the angular velocity ω is related to its linear velocity v by the equation ω = v / r, where r is the radius of the sphere.
Substituting the given values, we have:
Kinetic Energy = 1/2 * (2/5 * M * r^2) * (v / r)^2
The potential energy of the sphere at the bottom of the incline is given by:
Potential Energy = m * g * h
Where m is the mass of the sphere, g is the acceleration due to gravity, and h is the height of the incline.
Since the sphere rolls without slipping, the linear distance it travels up the incline is equal to the height of the incline. Therefore, the potential energy is also the change in gravitational potential energy, which is given by:
Potential Energy = m * g * h
Equating the initial mechanical energy to the final mechanical energy (0, when the sphere comes to a stop), we have:
1/2 * (2/5 * M * r^2) * (v / r)^2 + m * g * h = 0
Rearranging the equation, we can solve for h, the height of the incline:
h = - (1/2 * (2/5 * M * r^2) * (v / r)^2) / (m * g)
Substituting the given values:
h = - (1/2 * (2/5 * M * r^2) * (35 / r)^2) / (m * g)
Simplifying the equation will give you the final answer for the height h.