Respuesta :
I of the spool is (1/2)*m*r^2
The 5 kg mass puts a torque on the spool
torque = 5 kg*g*r
Angular acceleration, α: torque = I*α
So
5 kg*g*r = (1/2)*m*r^2 * α
Solve for α.
B. Rotation speed ω:
ω^2 = ωo^2 + 2*α*Θ
where ωo=0, α is from part B, Θ can be calculated from the length of the string and the circumference of the spool. (Note: Θ is in radians. 2π radians is one full revolution.)
The 5 kg mass puts a torque on the spool
torque = 5 kg*g*r
Angular acceleration, α: torque = I*α
So
5 kg*g*r = (1/2)*m*r^2 * α
Solve for α.
B. Rotation speed ω:
ω^2 = ωo^2 + 2*α*Θ
where ωo=0, α is from part B, Θ can be calculated from the length of the string and the circumference of the spool. (Note: Θ is in radians. 2π radians is one full revolution.)
The angular acceleration of the spool is [tex]51.28 \rm \ radians/s^2[/tex]. angular acceleration is the rate of change in the angular velocity of the object or body.
What angular acceleration?
It can be defined as the rate of change in the angular velocity of the object or body.
[tex]\alpha = \dfrac TI[/tex]
Where,
[tex]\alpha[/tex] - angular acceleration
[tex]\tau\\[/tex] - torque
[tex]I[/tex] - the moment of inertia
First, calculate the moment of inertia,
[tex]I = \dfrac 12 mr^2[/tex]
Where,
m - mass = 0.5 kg
r - radius = 0.075 m
Put the values in the formula,
[tex]I =\dfrac 12 (0.5 )(.075)^2\\\\I = 1.4 \times 10^{-3} \rm \ kg m^2[/tex]
Now, calculate the torque,
[tex]\tau = mgr[/tex]
Where,
[tex]g[/tex] - gravitational acceleration = 9.8 m/s²
So,
[tex]\tau = 5\times 9.8\times .075 \\\\\tau = 3.675 \rm \ Nm[/tex]
Now, calculate the angular acceleration,
[tex]\alpha = \dfrac {3.675}{ 1.4 \times 10^{-3}}\\\\\alpha = 51.28 \rm \ radians\s^2[/tex]
Therefore, the angular acceleration of the spool is [tex]51.28 \rm \ radians/s^2[/tex].
Learn more about angular acceleration:
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