Physics, 15.11.2019 04:31 lydiaa1

Case 1: a dj starts up her phonograph player. the turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get up to its full speed of f₁ = 78 revolutions per minute. case 2: the dj then changes the speed of the turntable from f₁ = 78 to f₂ = 120 revolutions per minute. she notices that the turntable rotates exactly n₂= 15 times while accelerating uniformly. t₁ = 11.9 secondsn₂ = 11 timespart (a) calculate the angular speed of the turntable while it is turning at f₁ = 78 in radians/second in case 1.part (b) how many revolutions does the turntable make while accelerating in case 1? part (c) calculate the magnitude of the angular acceleration of the turntable in case 1, in radians/second².part (d) calculate the magnitude of the angular acceleration of the turntable (in radians/second²) while increasing to 120 rpm (case 2).part (e) how long (in seconds) does it take for the turntable to go from f₁ = 78 to f₂ = 120 rpm?

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Answer from: FlayMaster101

This question involves the concept of the equations of motion for angular motion.

(a) Angular speed of the turntable is "8.17 rad/s" for case 1.

(b) The turntable makes "7.8 revs" revolutions in case 1.

(d) The magnitude of the angular acceleration of the turntable is "1.32 rad/s²" in case 2.

(e) It takes "3.33 s" for the turntable to go from 78 rpm to 120 rpm in case 2.

(a)

We will simply convert the angular frequency to angular speed by the following formula:

$\omega=2\pi f$

where.

ω = angular speed = ?

f = angular frequency = 78 rev/min

Therefore,

$\omega=(78\ rev/min)(\frac{2\pi\ rad}{1\ rev})(\frac{1\ min}{60\ s})$

ω = 8.17 rad/s

(c)

We will use the first equation of motion for the angular motion to find the angular acceleration in case 1:

$\omega_f=\omega_i+\alpha_1 t_1$

where,

$\omega_f$ = final angular speed = 8.17 rad/s

$\omega_i$ = initial angular speed = 0 rad/s

$\alpha_1$ = angular acceleration in case 1= ?

t₁ = time taken in case 1 = 11.9 s

Therefore,

$8.17\ rad/s=0\ rad/s+\alpha_1 (11.9\ s)\\\\\alpha_1=\frac{8.17\ rad/s}{11.9\ s}$

α₁ = 0.69 rad/s²

(b)

We will use the second equation of motion for the angular motion to find the no. of revolutions in case 1:

$\theta_1=\omega_it_1+\frac{1}{2}\alpha_1 t_1^2$

where,

θ₁ = no. of revolutions =?

$\omega_i$ = initial angular speed = 0 rad/s

$\alpha_1$ = angular acceleration in case 1= 0.69 rad/s²

t₁ = time taken in case 1 = 11.9 s

Therefore,

$\theta_1 = (0\ rev/s)(11.9\ s)+\frac{1}{2}(0.69\ rad/s^2)(11.9\ s)^2$

$\theta_1 = (48.85\ rad)(\frac{1\ rev}{2\pi\ rad})$

θ₁ = 7.8 revs

(d)

We will use the third equation of motion for the angular motion to find the angular acceleration in case 2:

$2\alpha_2 \theta_2 = \omega_f^2-\omega_i^2$

where,

$\omega_f$ = final angular speed = $(120\ rpm)(\frac{2\pi\ rad}{1\ rev})(\frac{1\ min}{60\ s}) = 12.57\ rad/s$

$\omega_i$ = initial angular speed = 8.17 rad/s

$\alpha_2$ = angular acceleration in case 2 = ?

$\theta_2$ = no. of revolutions in case 2 = $(15\ rev)(\frac{2\pi\ rad}{1\ rev}) = 94.25\ rad$

Therefore,

$2\alpha_2(94.25\ rad) = (12.57\ rad/s)^2-(8.17\ rad/s)^2\\\\\alpha_2=\frac{(12.57\ rad/s)^2-(8.17\ rad/s)^2}{2(94.25\ rad)}$

α₂ = 1.32 rad/s²

(e)

We will use the first equation of motion for the angular motion to find the time taken in case 2:

$\omega_f=\omega_i+\alpha_2 t_2$

where,

$\omega_f$ = final angular speed = 12.57 rad/s

$\omega_i$ = initial angular speed = 8.17 rad/s

t₂ = time taken in case 1 = ?

Therefore,

$12.57\ rad/s=8.17\ rad/s+()t_2\\\\t_2=\frac{12.57\ rad/s-8.17\ rad/s}{1.32\ rad/s^2}$

t₂ = 3.33 s

Learn more about the angular motion here:

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The attached picture shows the angular equations of motion.

Answer from: jeniferfayzieva2018

Part a)

$\omega = 8.17 rad/s$

Part b)

N = 7.74 rev

Part c)

$\alpha = 0.69 rad/s^2$

Part d)

$\alpha = 0.48 rad/s^2$

Part e)

t = 9.14 s

Explanation:

Part a)

Angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi(\frac{78}{60})$

$\omega = 8.17 rad/s$

Part b)

Since turn table is accelerating uniformly

so we will have

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{8.17 + 0}{2}(11.9)$

$2N\pi = 48.6$

N = 7.74 rev

Part c)

angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{8.17 - 0}{11.9}$

$\alpha = 0.69 rad/s^2$

Part d)

When its angular speed changes to 120 rpm

then we will have

$\omega_2 = 2\pi (\frac{120}{60})$

$\omega_2 = 12.56 rad/s$

number of turns revolved is 15 times

so we have

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

$12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)$

$\alpha = 0.48 rad/s^2$

Part e)

now for uniform acceleration we have

$\omega_f - \omega_i = \alpha t$

12.56 - 8.17 = 0.48 t

t = 9.14 s

Case 1: a dj starts up her phonograph player. the turntable accelerates uniformly from rest, and ta

Answer from: Quest

chemical energy --> gravitational potential energy --> kinetic energy --> elastic potential energy --> gravitational potential energy

explanation:

when emma starts the hike, the chemical energy stored in her body is used to climb up the mountain, and so she gains gravitational potential energy (because at the end, she will be at a higher height h above the ground). then she goes bungee jumping: as she falls, this gravitational potential energy is converted into kinetic energy (because her speed increases as she falls down), until the point when the rope stretches, and she stops moving. at that moment, all the kinetic energy is converted into elastic potential energy due to the stretching of the rope. then, this energy is released because the rope returns to its natural length, and emma gains a small increase in altitude $\delta h$ , so she gains some gravitational potential energy.