A disk with mass m = 9.8 kg and radius R = 0.31 m begins at rest and accelerates uniformly for t = 18.7 s, to a final angular speed of ω = 31 rad/s.
a. What is the angular acceleration of the disk?
b. What is the angular displacement over the 18.7 s?
c. What is the moment of inertia of the disk?
d. What is the change in rotational energy of the disk?
e. What is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed?
f. What is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed?
g. What is the final speed of a point on the disk half-way between the center of the disk and the rim?
h. What is the total distance a point on the rim of the disk travels during the 18.7 seconds?
is this a test ?
no it is HW
a. V = a*t = 31.
a*18.7 = 31,
a = 1.66 rad/s^2.
b. D = 0.5*a*t^2 = 0.5 * 1.66*18.7^2 = 290 rad.
d. KE = 0.5M*V^2 = 4.9*31^2 = Joules.
g. V = 31 rad/s.
h. Cir. = 2pi*r = 6.28 * 0.31 = 1.95 m.
d = 290rad * 1rev/6.28rad * 1.95m/rev = 90 m.
To solve these questions, we need to use the formulas and equations related to rotational kinematics and dynamics. Let's go step by step:
a. To find the angular acceleration (α), we can use the formula:
α = (ωf - ωi) / t
where ωf is the final angular velocity, ωi is the initial angular velocity (which is zero in this case since the disk starts at rest), and t is the time.
Plugging in the values given:
ωf = 31 rad/s,
ωi = 0 rad/s,
t = 18.7 s,
α = (31 - 0) / 18.7
α ≈ 1.658 rad/s²
Therefore, the angular acceleration of the disk is approximately 1.658 rad/s².
b. To find the angular displacement (θ), we can use the formula:
θ = ωi * t + 0.5 * α * t²
Since ωi = 0 rad/s, the equation becomes:
θ = 0.5 * α * t²
Plugging in the values given:
α = 1.658 rad/s²,
t = 18.7 s,
θ = 0.5 * 1.658 * (18.7)²
θ ≈ 301.295 radians
Therefore, the angular displacement over the 18.7 seconds is approximately 301.295 radians.
c. To find the moment of inertia (I) of the disk, we need to know the formula for the moment of inertia of a solid disk:
I = 0.5 * m * R²
Plugging in the values given:
m = 9.8 kg,
R = 0.31 m,
I = 0.5 * 9.8 * (0.31)²
I ≈ 0.451 kg·m²
Therefore, the moment of inertia of the disk is approximately 0.451 kg·m².
d. To find the change in rotational energy, we need to know the formula for rotational kinetic energy:
ΔE = 0.5 * I * (ωf - ωi)²
Since ωi = 0 rad/s, the equation becomes:
ΔE = 0.5 * I * ωf²
Plugging in the values given:
I = 0.451 kg·m²,
ωf = 31 rad/s,
ΔE = 0.5 * 0.451 * (31)²
ΔE ≈ 216.975 Joules
Therefore, the change in rotational energy of the disk is approximately 216.975 Joules.
e. To find the tangential component of the acceleration (at) of a point on the rim of the disk when it has accelerated to half its final angular speed, we can use the formula:
at = α * R
Plugging in the values given:
α = 1.658 rad/s²,
R = 0.31 m,
at = 1.658 * 0.31
at ≈ 0.513 m/s²
Therefore, the tangential component of the acceleration when the disk has accelerated to half its final angular speed is approximately 0.513 m/s².
f. To find the magnitude of the radial component of the acceleration (ar) of a point on the rim of the disk when it has accelerated to half its final angular speed, we can use the formula:
ar = R * ω²
Plugging in the values given:
R = 0.31 m,
ω = 0.5 * 31 rad/s (since it is half the final angular speed),
ar = 0.31 * (0.5 * 31)²
ar ≈ 7.645 m/s²
Therefore, the magnitude of the radial component of the acceleration when the disk has accelerated to half its final angular speed is approximately 7.645 m/s².
g. To find the final speed of a point on the disk halfway between the center and the rim, we need to know that the tangential speed is equal to the product of the radius and the angular speed:
v = R * ω
Plugging in the values given:
R = 0.31 m,
ω = 0.5 * 31 rad/s (since it is half the final angular speed),
v = 0.31 * (0.5 * 31)
v ≈ 4.795 m/s
Therefore, the final speed of a point on the disk halfway between the center and the rim is approximately 4.795 m/s.
h. To find the total distance a point on the rim of the disk travels during the 18.7 seconds, we need to know the formula relating angular displacement and radius:
s = θ * R
Plugging in the values given:
θ = 301.295 radians,
R = 0.31 m,
s = 301.295 * 0.31
s ≈ 93.388 meters
Therefore, the total distance a point on the rim of the disk travels during the 18.7 seconds is approximately 93.388 meters.