Answer :
Answer:
a) [tex]\alpha=0.5117\ rad.s^{-2}[/tex]
b) [tex]\theta=8.4\ rad[/tex]
Explanation:
Given:
- initial rotational speed of phonograph, [tex]\omega_i=0\ rad.s^{-1}[/tex]
- final rotational speed of phonograph, [tex]N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}[/tex]
- time taken for the acceleration, [tex]t=5.73\ s[/tex]
a)
Now angular acceleration:
[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]
[tex]\alpha=\frac{2.932-0}{5.73}[/tex]
[tex]\alpha=0.5117\ rad.s^{-2}[/tex]
b)
Using eq. of motion:
[tex]\theta=\omega_i.t+\frac{1}{2} \alpha.t^2[/tex]
[tex]\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2[/tex]
[tex]\theta=8.4\ rad[/tex]
a. The angular acceleration of the phonograph record is 0.51 [tex]rad/s^2[/tex]
b. The number of revolutions the phonograph record spun in this time interval is 8.37 revs.
Given the following data:
- Initial angular velocity = 0 rpm (since it accelerates from rest).
- Final angular velocity = 28 rpm
- Time = 5.73 seconds.
Conversion:
1 rpm = 0.1047 rad/s
28 rpm = X rad/s
Cross-multiplying, we have:
[tex]X = 28 \times 0.1047[/tex]
X = 2.93 rad/s
a. To find the angular acceleration of the phonograph record:
[tex]\alpha = \frac{w_f - w_i}{t}[/tex]
Where:
- [tex]w_f[/tex] is the final angular velocity.
- [tex]w_i[/tex] is the initial angular velocity.
- [tex]\alpha[/tex] is the angular acceleration.
- t is the time.
Substituting the given parameters into the formula, we have;
[tex]\alpha = \frac{2.93 - 0}{5.73}\\\\\alpha = \frac{2.93}{5.73}[/tex]
Angular acceleration = 0.51 [tex]rad/s^2[/tex]
b. To find how many revolutions it goes through in the process:
[tex]X = w_ot + \frac{1}{2} at^2\\\\X = 0(5.73) + \frac{1}{2} (0.51)(5.73)^2\\\\X = 0 + 0.255(32.8329)[/tex]
X = 8.37 revs.
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