A solid, uniform disk of radius 0.250 m and mass 60.4 kg rolls down a ramp of length 4.60 m that makes an angle of 18.0° with the horizontal. The disk starts from rest from the top of the ramp.
(a) Find the speed of the disk's center of mass when it reaches the bottom of the ramp.
m/s
(b) Find the angular speed of the disk at the bottom of the ramp.
rad/s
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To answer these questions, we can use the principles of conservation of energy and conservation of angular momentum.
(a) To find the speed of the disk's center of mass when it reaches the bottom of the ramp, we can use the principle of conservation of energy.
The potential energy at the top of the ramp is given by:
PE_top = m*g*h
where m is the mass of the disk (60.4 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height of the ramp.
The kinetic energy at the bottom of the ramp is given by:
KE_bottom = (1/2) * I * ω^2
where I is the moment of inertia of the disk and ω is the angular speed of the disk.
Since the disk starts from rest, the initial kinetic energy is zero.
According to the conservation of energy, the potential energy at the top of the ramp is equal to the kinetic energy at the bottom of the ramp:
m*g*h = (1/2) * I * ω^2
To find the moment of inertia (I) of the disk, we can use the formula for the moment of inertia of a solid disk rotating about its central axis:
I = (1/2) * m * r^2
where r is the radius of the disk (0.250 m).
Substituting this into the conservation of energy equation, we have:
m * g * h = (1/2) * ((1/2) * m * r^2) * ω^2
Simplifying the equation, we get:
g * h = (1/4) * r^2 * ω^2
Now we can solve for the angular speed (ω) at the bottom of the ramp:
ω = sqrt((4 * g * h) / r^2)
Substituting the given values, we have:
ω = sqrt((4 * 9.8 m/s^2 * 4.60 m) / (0.250 m)^2)
Calculating this, we find:
ω ≈ 14.35 rad/s
(b) To find the angular speed (ω) of the disk at the bottom of the ramp, we have just calculated it to be approximately 14.35 rad/s.