A 10.0-kg cylinder rolls without slipping on a rough surface. At the instant its center of amss has a speed of 10m/s, determine (a) the translational kinetic of its center of mass (b) the rotational kinetic energy about its center of mass and (c) its total kinetic energy.
can u check if this is the correct answer?
a) KE=.5Mv^2
KE=.5(10)(10)^2
KE=500J
B) KE=.25Mv^2
KE=.25(10)(10)^2
KE=250J
C) 250J+500J=750J
please tell me if that is correct ornot XD thanks
^please check that work i have shown above thanks
this is correct
According to my physics book with the answers in the back, this is correct.
cyndi used 1/4 because of this:
KErotational=1/2IW^2
=1/2(1/2mr^2)(V/r)^2
=1/2(1/2mr^2)(V^2/r^2)
=1/4mV^2
as you can see the radii cancel and the 1/2's multiply together to make 1/4
why are you using .25 for b??
Well, let's see if your calculations are correct.
(a) To find the translational kinetic energy of the center of mass, you have to use the formula KE = 0.5mv^2. Plugging in the values:
KE = 0.5(10)(10)^2
KE = 0.5(10)(100)
KE = 500 J
So, you got the correct answer for part (a). Good job!
(b) Now, for the rotational kinetic energy about the center of mass, the formula is KE = 0.5Iω^2, where I is the moment of inertia and ω is the angular velocity. Since the cylinder is rolling without slipping, we can use the relationship between linear and angular velocity: v = rω, where r is the radius of the cylinder.
So, ω = v/r = 10/1 = 10 rad/s.
The moment of inertia for a solid cylinder rotating about its central axis is I = 0.5MR^2, where M is the mass and R is the radius. Plugging in the values:
KE = 0.5(10)(1^2)(10^2)
KE = 0.5(10)(1)(100)
KE = 500 J
Uh-oh, it seems there's a mistake in your calculation for part (b). The correct answer should be 500 J, not 250 J. Nice try, though!
(c) Finally, to find the total kinetic energy, you simply sum up the translational and rotational kinetic energies:
KE_total = KE_translational + KE_rotational
KE_total = 500 J + 500 J
KE_total = 1000 J
Oops, there's another mistake in your calculation for part (c). The correct answer is 1000 J, not 750 J.
So, the corrected answers to your questions are:
(a) The translational kinetic energy of the center of mass is 500 J.
(b) The rotational kinetic energy about the center of mass is 500 J.
(c) The total kinetic energy is 1000 J.
Hope that clears things up for you!
To determine the translational and rotational kinetic energy of the cylinder and its total kinetic energy, we can use the following formulas:
(a) Translational Kinetic Energy:
The translational kinetic energy of an object is given by the formula: KE_trans = 0.5 * m * v^2
Here, m represents the mass of the cylinder, and v represents the speed of its center of mass.
So, substituting the given values:
m = 10.0 kg
v = 10 m/s
KE_trans = 0.5 * 10.0 kg * (10 m/s)^2
KE_trans = 0.5 * 10.0 kg * 100 m^2/s^2
KE_trans = 500 J
Therefore, the correct answer for the translational kinetic energy is 500 J.
(b) Rotational Kinetic Energy about the Center of Mass:
The rotational kinetic energy about the center of mass of a rolling object is given by the formula: KE_rot = 0.5 * I * ω^2
Here, I represents the moment of inertia of the cylinder and ω represents its angular velocity.
For a cylinder rolling without slipping, the moment of inertia about its center of mass is given by: I = 0.5 * m * r^2
Where r represents the radius of the cylinder.
Since the radius is not provided, let's assume it is known or can be calculated separately.
Assuming r = 1 (unit), we can substitute the given values:
m = 10.0 kg
r = 1 m (assumption)
ω = v / r = 10 m/s / 1 m = 10 rad/s
I = 0.5 * 10.0 kg * (1 m)^2
I = 0.5 * 10.0 kg * 1 m^2
I = 5 kg * m^2
KE_rot = 0.5 * 5 kg * m^2 * (10 rad/s)^2
KE_rot = 0.5 * 5 kg * m^2 * 100 rad^2/s^2
KE_rot = 250 J
Therefore, the correct answer for the rotational kinetic energy about the center of mass is 250 J.
(c) Total Kinetic Energy:
The total kinetic energy is the sum of the translational and rotational kinetic energies.
Total KE = KE_trans + KE_rot
Total KE = 500 J + 250 J
Total KE = 750 J
Hence, your answer is correct. The total kinetic energy is indeed 750 J.