a disk i = 0.7 kgm and radius 60 cm is rotating a 12 rpm. there is an external torque applied to the outer edge and the rotational speed increases to 35 rpm in 8 sec.
angular acceleration?
angle the disk rotates?
magnitude of applied force (assume tangential)?
To find the angular acceleration of the disk, we can use the formula:
angular acceleration (α) = (final angular speed - initial angular speed) / time
Final angular speed = 35 rpm
Initial angular speed = 12 rpm
Time = 8 sec
Converting rpm to rad/s:
1 rpm = 2π rad/min
35 rpm = 35 * 2π rad/min = 70π rad/s
12 rpm = 12 * 2π rad/min = 24π rad/s
Plugging in the values, we get:
α = (70π - 24π) / 8
= 46π / 8
= 23π / 4
Therefore, the angular acceleration is 23π / 4 rad/s².
To find the angle the disk rotates, we can use the formula:
θ = ω * t + 0.5 * α * t²
ω = initial angular speed = 12 rpm = 24π rad/s
t = time = 8 sec
α = angular acceleration = 23π / 4 rad/s²
Plugging in the values, we get:
θ = (24π * 8) + (0.5 * (23π / 4) * (8^2))
= 192π + 0.5 * 23π * 64
= 192π + 23π * 32
= 192π + 736π
= 928π
Therefore, the angle the disk rotates is 928π radians.
To find the magnitude of the applied force (assume tangential), we can use the formula:
Torque = moment of inertia * angular acceleration
Moment of inertia (I) can be calculated using the formula:
Moment of inertia (I) = 0.5 * mass * radius²
Mass (m) = 0.7 kg
Radius (r) = 60 cm = 0.6 m
Angular acceleration (α) = 23π / 4 rad/s² (from earlier calculation)
Plugging in the values, we get:
I = 0.5 * 0.7 * (0.6^2)
= 0.5 * 0.7 * 0.36
= 0.126 kgm²
Torque = 0.126 * (23π / 4)
= 2.079π
Force (F) = Torque / radius
= (2.079π) / 0.6
= 3.465π
Therefore, the magnitude of the applied force (assuming tangential) is approximately 3.465π units.
To find the angular acceleration, angle the disk rotates, and magnitude of the applied force, we can use the following formulas derived from the laws of rotational motion:
1. Angular acceleration (α):
α = Δω / Δt
Where:
Δω = change in angular velocity (final angular velocity - initial angular velocity)
Δt = change in time (final time - initial time)
2. Angle rotated (θ):
θ = ω_i * t + (1/2) * α * t^2
Where:
ω_i = initial angular velocity
t = time
3. Magnitude of applied force (F):
F = I * α
Where:
I = moment of inertia of the disk
α = angular acceleration
Now, let's calculate the values:
Given:
Mass (m) = 0.7 kg
Radius (r) = 60 cm = 0.6 m
Initial angular velocity (ω_i) = 12 rpm * (2π rad/1 min) * (1 min/60 sec) = 2π/5 rad/sec
Final angular velocity (ω_f) = 35 rpm * (2π rad/1 min) * (1 min/60 sec) = 7π/6 rad/sec
Change in time (Δt) = 8 seconds
First, let's find the angular acceleration (α):
Δω = ω_f - ω_i = (7π/6) - (2π/5) = (35π - 12π) / 30 = 23π / 30 rad/sec
α = Δω / Δt = (23π/30) / 8 ≈ 0.3038 rad/sec^2
Next, let's find the angle rotated (θ):
θ = ω_i * t + (1/2) * α * t^2 = (2π/5) * 8 + (1/2) * (0.3038) * (8^2) ≈ 10.086 rad
Lastly, let's find the magnitude of the applied force (F):
The moment of inertia (I) of a disk is given by I = (1/2) * m * r^2
I = (1/2) * (0.7) * (0.6)^2 = 0.126 kgm^2
F = I * α = 0.126 * 0.3038 ≈ 0.0383 N
Therefore, the angular acceleration is approximately 0.3038 rad/sec^2, the angle rotated is approximately 10.086 radians, and the magnitude of the applied force is approximately 0.0383 N.