A uniform solid cylinder of radius R and a thin uniform spherical shell of radius R both roll without slipping. If both objects have the same mass and the same kinetic energy, what is the ratio of the linear speed of the cylinder to the linear speed of the spherical shell?
To start solving this problem, we can use the concept of rotational kinetic energy and linear kinetic energy.
1. Let's begin by defining the variables:
- R: radius of the cylinder and spherical shell
- m: mass of the cylinder and spherical shell
- v_cylinder: linear speed of the cylinder
- v_shell: linear speed of the spherical shell
2. The rotational kinetic energy of the cylinder is given by:
K_rot_cylinder = (1/2) * I_cylinder * ω_cylinder^2
Since the cylinder is rolling without slipping, its angular velocity ω_cylinder can be related to its linear speed v_cylinder:
ω_cylinder = v_cylinder / R
The moment of inertia of a solid cylinder about its central axis is given by:
I_cylinder = (1/2) * m * R^2
Substituting these values into the equation for rotational kinetic energy of the cylinder:
K_rot_cylinder = (1/2) * (1/2) * m * R^2 * (v_cylinder / R)^2
= (1/4) * m * v_cylinder^2
3. Similarly, the rotational kinetic energy of the spherical shell is given by:
K_rot_shell = (1/2) * I_shell * ω_shell^2
Since the spherical shell is rolling without slipping, its angular velocity ω_shell can be related to its linear speed v_shell:
ω_shell = v_shell / R
The moment of inertia of a thin spherical shell about its central axis is given by:
I_shell = (2/3) * m * R^2
Substituting these values into the equation for rotational kinetic energy of the shell:
K_rot_shell = (1/2) * (2/3) * m * R^2 * (v_shell / R)^2
= (1/3) * m * v_shell^2
4. The total kinetic energy of the cylinder is the sum of its linear kinetic energy and rotational kinetic energy:
K_total_cylinder = (1/2) * m * v_cylinder^2 + (1/4) * m * v_cylinder^2
= (3/4) * m * v_cylinder^2
5. The total kinetic energy of the spherical shell is the sum of its linear kinetic energy and rotational kinetic energy:
K_total_shell = (1/2) * m * v_shell^2 + (1/3) * m * v_shell^2
= (5/6) * m * v_shell^2
6. Given that both objects have the same mass and the same total kinetic energy, we can equate K_total_cylinder and K_total_shell:
(3/4) * m * v_cylinder^2 = (5/6) * m * v_shell^2
7. Dividing both sides of the equation by m and simplifying:
(3/4) * v_cylinder^2 = (5/6) * v_shell^2
8. Taking the square root of both sides to solve for the ratio of their linear speeds:
v_cylinder / v_shell = sqrt[(5/6) / (3/4)]
= sqrt[(5/6) * (4/3)]
= sqrt[(20/18)]
= sqrt[(10/9)]
Therefore, the ratio of the linear speed of the cylinder to the linear speed of the spherical shell is sqrt(10/9).
To find the ratio of the linear speed of the cylinder to the linear speed of the spherical shell, we need to analyze the motion of both objects using the principle of conservation of energy and the conditions of rolling without slipping.
Let's start with the cylinder. Since it is rolling without slipping, we can consider the kinetic energy of both the rotational and translational motion. The total kinetic energy (Kc) of the cylinder can be written as the sum of the translational kinetic energy (Kt) and the rotational kinetic energy (Kr):
Kc = Kt + Kr
The translational kinetic energy is given by:
Kt = (1/2) * m * Vc^2
where m is the mass of the cylinder and Vc is the linear speed (velocity) of the cylinder.
The rotational kinetic energy is given by:
Kr = (1/2) * Ic * ω^2
where Ic is the moment of inertia of the cylinder about its axis of rotation and ω is the angular speed of the cylinder.
Similarly, for the spherical shell, the total kinetic energy (Ks) will also be the sum of the translational kinetic energy (Kt) and the rotational kinetic energy (Kr). However, since the shell is a hollow object, its moment of inertia (Is) will be different from that of the solid cylinder:
Ks = Kt + Kr
Kt = (1/2) * m * Vs^2
Kr = (1/2) * Is * ω^2
We are given that the mass and the kinetic energy of the two objects are the same. So we can equate the total kinetic energies:
Kc = Ks
(1/2) * m * Vc^2 + (1/2) * Ic * ω^2 = (1/2) * m * Vs^2 + (1/2) * Is * ω^2
Now, if we make use of the relationship between linear speed (V) and angular speed (ω) for rolling motion, which is given by:
V = R * ω
where R is the radius of the object (both for the cylinder and the shell), we can substitute Vc = R * ωc and Vs = R * ωs into the equation above. This gives us:
(1/2) * m * (R * ωc)^2 + (1/2) * Ic * ωc^2 = (1/2) * m * (R * ωs)^2 + (1/2) * Is * ωs^2
Simplifying and canceling out common factors:
(1/2) * m * R^2 * ωc^2 + (1/2) * Ic * ωc^2 = (1/2) * m * R^2 * ωs^2 + (1/2) * Is * ωs^2
Notice that R^2 is common to both sides of the equation. Rearranging and simplifying further:
m * R^2 * ωc^2 + Ic * ωc^2 = m * R^2 * ωs^2 + Is * ωs^2
We know that the moment of inertia for a solid cylinder about its axis of rotation (Ic) and the moment of inertia for a thin spherical shell about its axis of rotation (Is) are given by the formulas:
Ic = (1/2) * m * R^2
Is = (2/3) * m * R^2
Substituting these values into the equation:
m * R^2 * ωc^2 + (1/2) * m * R^2 * ωc^2 = m * R^2 * ωs^2 + (2/3) * m * R^2 * ωs^2
Canceling out the common factor (m * R^2) and simplifying:
ωc^2 + (1/2) * ωc^2 = ωs^2 + (2/3) * ωs^2
Simplifying further:
(3/2) * ωc^2 = (5/3) * ωs^2
Now, since V = R * ω, we can solve for the ratio of linear speeds (Vc/Vs):
Vc / Vs = (R * ωc) / (R * ωs) = ωc / ωs
Substituting the above equation relating angular speeds into this ratio:
Vc / Vs = (3/2) * ωc^2 / (5/3) * ωs^2
Vc / Vs = (9/10) * (ωc^2 / ωs^2)
Finally, recall that the total kinetic energy of both objects is equal, therefore:
(1/2) * m * Vc^2 = (1/2) * m * Vs^2
Canceling out the common factor (1/2 * m) and rearranging:
Vc^2 = Vs^2
This means ωc^2 / ωs^2 = 1, and hence:
Vc / Vs = (9/10) * 1
Therefore, the ratio of the linear speed of the cylinder to the linear speed of the spherical shell is 9/10.
K.E. = Etranslational + Erotational
For either cylinder or sphere, V = R*w
For a uniform solid sphere:
K.E. = (1/2)MV^2 + (1/2)I w^2
= (1/2)MV^2 + (1/2)*(2/5)M*R^2*(V/R)^2
= (7/10) M V^2
For a uniform solid cylinder:
K.E. = (1/2)MV^2 + (1/2)((1/2)MR^2*(V/w)^2
= (3/4)M V^2
If masses are equal, AND KE's are equal,
V(cylinder)^2/V(sphere)^2 = (4/3)/(10/7)
= 0.93333
V(cylinder)/V(sphere) = 0.9661