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Roman55 [17]
3 years ago
10

The horse on a carousel is 3.5m from the central axis.A. If the carousel rotates at 0.13 rev/s , how long does it take the horse

to go around twice?B. How fast is a child on the horse going (in m/s)?
Physics
1 answer:
Kitty [74]3 years ago
8 0

Answer:

a. 15.4 seconds

b. 0.455 m/s

Explanation:

a. The carousel rotates at 0.13 rev/s.

This means that it takes the carousel 1 sec to make 0.13 of an entire revolution.

This means that time it will take to make a complete revolution is:

1 / 0.13 = 7.7 seconds

Therefore, the time it will take to make 2 revolutions is:

2 * 7.7 = 15.4 seconds

b. Let us calculate the linear velocity. Angular velocity is given as:

\omega = v / r

where v = linear velocity and r = radius

The radius of the circle is 3.5 m and the angular velocity is 0.13 rev/s, therefore:

0.13 = v / 3.5

v = 3.5 * 0.13 = 0.455 m/s

Linear velocity is 0.455 m/s

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it
Degger [83]

Answer:

a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

d)\alpha_2=4.1454\,rad.s^{-2}

e)t_2=1.061\,s

Explanation:

Given that:

  • initial speed of turntable, N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

4 0
3 years ago
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