A uniform disk with mass 41.0 kgkg and radius 0.220 mm is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 29.5 NN is applied tangent to the rim of the disk.Part AWhat is the magnitude v𝑣 of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.190 revolution?Express your answer with the appropriate units.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typev𝑣 =ratnothing
Question
A uniform disk with mass 41.0 kgkg and radius 0.220 mm is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 29.5 NN is applied tangent to the rim of the disk.Part AWhat is the magnitude v𝑣 of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.190 revolution?Express your answer with the appropriate units.Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value typev𝑣 =ratnothing
Solution
To solve this problem, we need to use the principles of rotational motion.
Step 1: Calculate the torque (τ) produced by the force. Torque is given by the formula τ = F*r, where F is the force and r is the radius of the disk.
τ = 29.5 N * 0.220 m = 6.49 Nm
Step 2: Calculate the angular acceleration (α) using the formula τ = Iα, where I is the moment of inertia of the disk. The moment of inertia of a disk is given by I = 0.5m*r^2, where m is the mass of the disk and r is its radius.
I = 0.5 * 41.0 kg * (0.220 m)^2 = 0.99 kg*m^2
So, α = τ / I = 6.49 Nm / 0.99 kg*m^2 = 6.56 rad/s^2
Step 3: Calculate the angular displacement (θ) for 0.190 revolution. 1 revolution is equal to 2π radians, so θ = 0.190 * 2π = 1.19 rad
Step 4: Use the formula for final angular velocity (ω) in terms of initial angular velocity (ω0), angular acceleration (α), and angular displacement (θ): ω^2 = ω0^2 + 2αθ. The disk starts from rest, so ω0 = 0.
ω = sqrt(2 * α * θ) = sqrt(2 * 6.56 rad/s^2 * 1.19 rad) = 4.81 rad/s
Step 5: The tangential velocity (v) of a point on the rim of the disk is given by v = ω*r.
v = 4.81 rad/s * 0.220 m = 1.06 m/s
So, the magnitude of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.190 revolution is 1.06 m/s.
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