Answer:2.7 joules
Step-by-step explanation:
Rolling Cylinder Kinetic Energy.
A cylindrical solid object of mass 400g and radius of 20cm is rolling down as inclined plane at speed of 3m/s. calculate the kinetic energy of the given object.
To calculate the kinetic energy of a rolling cylinder, we need to use the formula for rotational kinetic energy, which is:
K_rotational = (1/2) * I * ω^2
where I is the moment of inertia and ω is the angular velocity.
For a cylinder rolling without slipping down an inclined plane, the total kinetic energy is the sum of translational kinetic energy and rotational kinetic energy, which is:
K_total = K_translational + K_rotational
where K_translational is given by:
K_translational = (1/2) * m * v^2
where m is the mass of the cylinder, and v is its linear velocity.
First, let's find the moment of inertia of the cylinder. For a solid cylinder, the moment of inertia is given by:
I = (1/2) * m * r^2
where r is the radius of the cylinder. Substituting the given values, we get:
I = (1/2) * 0.4 kg * (0.2 m)^2 = 0.008 kg·m^2
Next, let's find the linear velocity of the cylinder. The speed of the cylinder is given as 3 m/s, so the linear velocity is:
v = 3 m/s
Now we can calculate the translational kinetic energy:
K_translational = (1/2) * m * v^2 = (1/2) * 0.4 kg * (3 m/s)^2 = 0.9 J
Finally, we can calculate the rotational kinetic energy:
K_rotational = (1/2) * I * ω^2
Since the cylinder is rolling without slipping, the linear velocity is related to the angular velocity as:
v = ω * r
Rearranging this equation, we get:
ω = v / r = 3 m/s / 0.2 m = 15 rad/s
Substituting the values, we get:
K_rotational = (1/2) * 0.008 kg·m^2 * (15 rad/s)^2 = 1.8 J
Therefore, the total kinetic energy of the cylinder is:
K_total = K_translational + K_rotational = 0.9 J + 1.8 J = 2.7 J
Therefore, the kinetic energy of the given object is 2.7 Joules.