Respuesta :
Answer:
a) [tex]\omega = 0.421\,\frac{rev}{s}[/tex], b) [tex]\Delta \theta = 0.066\,rev[/tex], c) [tex]v = 0.966\,\frac{mm}{s}[/tex], d) [tex]a = 3.293\,\frac{mm}{s^{2}}[/tex]
Explanation:
a) The angular velocity of the turntable after [tex]0.200\,s[/tex].
[tex]\omega = \omega_{o} + \alpha\cdot \Delta t[/tex]
[tex]\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)[/tex]
[tex]\omega = 0.421\,\frac{rev}{s}[/tex]
b) The change in angular position is:
[tex]\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}[/tex]
[tex]\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}[/tex]
[tex]\Delta \theta = 0.066\,rev[/tex]
c) The tangential speed of a point on the rim of the turn-table:
[tex]v = r\cdot \omega[/tex]
[tex]v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )[/tex]
[tex]v = 9.655\times 10^{-4}\,\frac{m}{s}[/tex]
[tex]v = 0.966\,\frac{mm}{s}[/tex]
d) The tangential and normal components of the acceleration of the turn-table:
[tex]a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )[/tex]
[tex]a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}[/tex]
[tex]a_{t} = 2.078\,\frac{mm}{s}[/tex]
[tex]a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}[/tex]
[tex]a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}[/tex]
[tex]a_{n} = 2.554\,\frac{mm}{s^{2}}[/tex]
The magnitude of the resultant acceleration is:
[tex]a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}[/tex]
[tex]a = 3.293\,\frac{mm}{s^{2}}[/tex]
For the electric turntable 0.730 mm in diameter that is rotating with an initial angular velocity of 0.240 rev/s and a constant angular acceleration of 0.906 rev/s², we have:
a) The angular velocity of the turntable after 0.200 s is 2.65 rad/s.
b) The number of revolutions in 0.200 s is 0.066.
c) The tangential speed of a point on the rim of the turntable at a time of 0.200 s is 9.67x10⁻⁴ m/s.
d) The magnitude of the resultant acceleration of a point on the rim at t = 0.200 s is 3.30x10⁻³ m/s².
a) We can calculate the angular velocity with the following equation:
[tex] \omega_{f} = \omega_{i} + \alpha t [/tex] (1)
Where:
[tex] \omega_{f} [/tex]: is the final angular velocity =?
[tex] \omega_{i} [/tex]: is the initial angular velocity = 0.240 rev/s
α: is the angular acceleration = 0.906 rev/s²
t: is the time = 0.200 s
By entering the above values into equation (1), we have:
[tex] \omega_{f} = 0.240 \frac{rev}{s}*\frac{2\pi rad}{1 rev} + 0.906 \frac{rev}{s^{2}}*\frac{2\pi rad}{1 rev}*0.200 s = 2.65 rad/s [/tex]
Hence, the angular velocity of the turntable after 0.200 s is 2.65 rad/s.
b) We can find the number of revolutions as follows
[tex] \theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha t^{2} [/tex] (2)
Where:
[tex] \theta_{f} [/tex]: is the final displacement =?
[tex] \theta_{i} [/tex]: is the initial displacement = 0
Then, the number of revolutions is (eq 2):
[tex] \theta_{f} = 0 + 0.240 \frac{rev}{s}*0.200 s + \frac{1}{2}0.906 \frac{rev}{s^{2}}*(0.200 s)^{2} = 0.066 rev[/tex]
c) The tangential speed is equal to:
[tex] v = \omega_{f}*r [/tex] (3)
Where:
r: is the radius of the turntable = d/2 = 0.730/2 mm = 0.365 mm
Hence, the tangential speed is (eq 3):
[tex] v = 2.65 rad/s*0.365 \cdot 10^{-3} m = 9.67 \cdot 10^{-4} m/s [/tex]
d) The magnitude of the resultant acceleration is equal to the sum of the centripetal and tangential acceleration:
[tex] a_{r} = \sqrt{a_{t}^{2} + a_{c}^{2}} [/tex] (4)
Where:
[tex]a_{c}[/tex]: is the centripetal acceleration = [tex]\omega_{f}^{2}*r[/tex]
[tex]a_{t}[/tex]: is the tangential acceleration = α*r
So, the resultant acceleration is (eq 4):
[tex] a_{r} = \sqrt{(\alpha*r)^{2} + (\omega_{f}^{2}*r)^{2}} [/tex]
[tex] a_{r} = \sqrt{(0.906 \frac{rev}{s^{2}}*\frac{2\pi rad}{1 rev}*0.365 \cdot 10^{-3} m)^{2} + ((2.65 rad/s)^{2}*0.365 \cdot 10^{-3} m)^{2}} = 3.30 \cdot 10^{-3} m/s^{2} [/tex]
Therefore, the magnitude of the resultant acceleration is 3.30x10⁻³ m/s².
Learn more about circular motion here:
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I hope it helps you!
