A solid uniform marble and a block of ice, each with the same mass, start from rest at the same height above the bottom of a hill and move down it. The marble rolls without slipping, but the ice slides without friction.
Find the speed of each of these objects when it reaches the bottom of the hill.
V_ice=
V_marble=
For ice, you can use conservation of energy. If h is the height that the block descends,
(1/2)MV^2 = M g h
V = sqrt (2gH)
For the marble block, since it rolls, I assume it is a cylinder. Static friction is what helps it roll instead of slip, but it does not result in a of of energy, since there is no relative motion at the line of contact. Equate the potential energy change to the increase in both translational and rotational kinetic energy. V = r w, where w is the angular velocity of the rolling cyliner. The moment of inertia is I = (1/2)M r^2
M g H = (1/2) M V^2 + (1/2)I w^2
= (1/2) M V^2 + (1/2)(1/2)Mr^2(V/r)^2
= (3/4) M V^2
V = sqrt[(4/3)gH]
actually for the second part, it wouldn't be a cylinder but a sphere. thus beta = 2/5. the velocity equation in this case would be V = sqrt[(2gH)/(1+beta)]
To find the speed of each object when it reaches the bottom of the hill, we can use the principle of conservation of energy.
Let's assume that the height of the hill is denoted by h, and the mass of each object is denoted by m.
1. For the marble:
The marble rolls without slipping, which means it has both rotational and translational kinetic energy. The total mechanical energy of the marble is given by the sum of its translational kinetic energy and rotational kinetic energy.
Translational kinetic energy (TKE) = (1/2)mv^2 (where v is the velocity of the marble)
Rotational kinetic energy (RKE) = (1/2)Iω^2 (where I is the moment of inertia of the marble and ω is its angular velocity)
Since the marble is rolling without slipping, the velocity v and the angular velocity ω are related by v = Rω, where R is the radius of the marble.
The total mechanical energy (E_total) of the marble is the sum of its TKE and RKE:
E_total = TKE + RKE
E_total = (1/2)mv^2 + (1/2)Iω^2
E_total = (1/2)mv^2 + (1/2)(2/5)mr^2ω^2 (for a solid uniform marble, I = (2/5)mr^2)
Now, let's consider the potential energy (PE) of the marble at the top of the hill. When the marble reaches the bottom of the hill, all of its potential energy will be converted to kinetic energy.
PE = mgh (where g is the acceleration due to gravity)
Equating the total mechanical energy to the potential energy:
E_total = PE
(1/2)mv^2 + (1/2)(2/5)mr^2ω^2 = mgh
However, we know that v = Rω. Substituting this expression into the equation above, we get:
(1/2)mv^2 + (1/2)(2/5)mr^2(v/R)^2 = mgh
Simplifying the equation:
(1/2)mv^2 + (1/2)(2/5)mr^2(v^2/R^2) = mgh
(1/2)mv^2 + (1/2)(2/5)mr^2(v^2/R^2) - mgh = 0
This equation is quadratic in v^2, so we can solve it using the quadratic formula. However, we know that the marble will pick up speed as it rolls down the hill, so the positive solution will be the correct one.
Using the quadratic formula, we get:
v^2 = (-b + sqrt(b^2 - 4ac)) / 2a
where a = (1/2)m + (1/2)(2/5)mr^2/R^2
b = 0
c = -mgh
Simplifying the equation and taking the positive solution:
v^2 = sqrt(4(1/2)mgh) / (2[(1/2)m + (1/2)(2/5)mr^2/R^2])
v = sqrt(2gh / (1 + (2/5)(r^2/R^2)))
2. For the ice block:
Since the ice block slides without friction, it does not have any rotational kinetic energy. Therefore, the total mechanical energy of the ice block is equal to its translational kinetic energy.
Using the same equation for potential energy as before:
PE = mgh
Equating the total mechanical energy to the potential energy:
E_total = PE
(1/2)mv_ice^2 = mgh
Simplifying the equation:
v_ice^2 = 2gh
v_ice = sqrt(2gh)
So, the speed of each object when it reaches the bottom of the hill is:
V_ice = sqrt(2gh)
V_marble = sqrt(2gh / (1 + (2/5)(r^2/R^2)))
To find the speed of each object when they reach the bottom of the hill, we can use the principle of conservation of mechanical energy. This principle states that the initial mechanical energy of a system is equal to its final mechanical energy if there are no external forces acting on it.
For the solid uniform marble, its mechanical energy is given by the sum of its kinetic energy (KE) and its gravitational potential energy (PE). Since the marble rolls without slipping, its kinetic energy can be expressed as translational kinetic energy (1/2mv^2) and rotational kinetic energy (1/2Iω^2), where m is the mass of the marble, v is its linear velocity, I is its moment of inertia, and ω is its angular velocity.
For the block of ice, since it slides without friction, it only has kinetic energy (KE) due to its linear velocity (v).
The gravitational potential energy of both objects can be calculated using the equation PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height from which the object is released.
To find the speed of each object at the bottom of the hill, we can equate their initial mechanical energy to their final mechanical energy.
For the solid uniform marble:
Initial mechanical energy = mgh
Final mechanical energy = KE (translational) + KE (rotational) + PE
For the block of ice:
Initial mechanical energy = mgh
Final mechanical energy = KE
Since both objects have the same initial mechanical energy (as they start from the same height), we can set their final mechanical energies equal to each other:
mgh = (1/2mv^2 + 1/2Iω^2) + KE
Now, we need to consider the relationship between linear velocity and angular velocity for the rolling marble. For a solid uniform sphere, the moment of inertia (I) can be expressed as I = (2/5)mr^2, where r is the radius of the sphere.
We can substitute the moment of inertia into the equation and simplify:
mgh = (1/2mv^2 + 1/2(2/5)mr^2ω^2) + KE
Since the marble rolls without slipping, its linear velocity (v) can be related to its angular velocity (ω) by the equation v = ωr. Substituting this relationship into the equation:
mgh = (1/2mv^2 + 1/2(2/5)mr^2(v/r)^2) + KE
Simplifying further:
mgh = (1/2mv^2 + (1/5)mv^2) + KE
mgh = (7/10)mv^2 + KE
Now, we can solve for the speed of the marble (v_marble):
v_marble = sqrt((2/7)(gh))
For the block of ice, since it slides without friction, its final mechanical energy is just its kinetic energy:
mgh = KE
Solving for the speed of the block of ice (v_ice):
v_ice = sqrt(gh)
Therefore, the speed of the ice block (v_ice) when it reaches the bottom of the hill is equal to the square root of gh, while the speed of the solid uniform marble (v_marble) is equal to the square root of (2/7)(gh).
Please note that these equations assume an idealized scenario, neglecting any external forces, air resistance, or other factors that may affect the motion of the objects.