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A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.3 m and moment of inertia
820kg⋅m2. The platform rotates without friction with an angular velocity of 0.85 rad/s. The person walks radially to the edge of the platform.
(a) Calculate the angular velocity when the person reaches the edge.
(b) Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk

Respuesta :

opudodennis

Answer:

(a) 0.426 rad/s

(b) Before 296.225 J and After 148.5 J

Explanation:

At the center of merry-go-round, the person's weight is 75 Kg with moment of inertia of 820 kgm2 and the initial angular velocity is 0.85 rad/s

At the edge, the new angular velocity gained depends on the new moment of inertia

Here, the final moment of inertia is given by Initial moment of inertia + [tex]MR^{2}[/tex] where M is the mass and R is the radius of merry-go-round

Final moment of inertia=[tex]820+(75*3.3^{2})=1636.75[/tex]

From the law of conservation of angular momentum as torque is zero

[tex]I_i\times\omega_1=I_f\times\omega_2[/tex]  and making [tex]\omega_2[/tex] the subject then

[tex]\omega_2=\frac {I_i\times\omega}{I_f}[/tex]

where [tex]I_i[/tex] and [tex]I_f[/tex] are initial and final moment of inertia, [tex]\omega_1[/tex] and [tex]\omega_2[/tex] are initial and final angular velocity.

Substituting the provided values

[tex]\omega_2=\frac {0.85\times 820}{1636.75}=0.425843898\approx 0.426 rad/s[/tex]

(b)

Initial rotational kinetic energy is given by

[tex]0.5I_i\times \omega_1^{2}[/tex]

[tex]KE_i=0.5\times 820\times 0.85^{2}=296.225 J[/tex]

The final rotational kinetic energy is given by

[tex]0.5I_f\times \omega_2^{2}[/tex]

[tex]KE_i=0.5\times 1636.75\times 0.426^{2}\approx 148.5 J[/tex]

Lanuel

a. The angular velocity when the person reaches the edge is 0.42 rad/s.

b. The rotational kinetic energy of the system of platform plus person before and after the person's walk are 296.23 Joules and 140.44 Joules respectively.

Given the following data:

  • Mass of person = 75 kg
  • Initial radius = 3.3 m
  • Initial angular speed = 0.85 rad/s
  • Moment of inertia (person + platform) = 820 [tex]kgm^2[/tex]  

a. To calculate the angular velocity when the person reaches the edge, we would use the law of conservation of angular momentum:

Applying the law of conservation of angular momentum:

[tex]I_i\omega_i = I_f\omega_f[/tex]  ....equation 1.

Where:

  • [tex]I_i[/tex] is the initial moment of inertia.
  • [tex]\omega_i[/tex] is the initial angular speed.
  • [tex]I_f[/tex] is the final moment of inertia.
  • [tex]\omega_f[/tex] is the final angular speed.

First of all, we would determine both the initial and final angular momentum of the person and the platform.

Note: The initial angular momentum of the person would be zero (0).

For initial angular momentum:

[tex]I_i\omega_i = 0+820 \times 0.85\\\\I_i\omega_i = 697\;kgm^2/s[/tex]  ....equation 2.

For final angular momentum:

[tex]I_f\omega_f = 75(3.3)^2\omega_f + 820\omega_f \\\\I_f\omega_f = 75(10.89)\omega_f + 820\omega_f\\\\I_f\omega_f = 816.75\omega_f + 820\omega_f\\\\I_f\omega_f = 1636.75\omega_f \;kgm^2/s[/tex].....equation 3.

Equating both eqn 2 and eqn 3, we have:

[tex]679 = 1636.75\omega_f\\\\\omega_f = \frac{679}{1636.75} \\\\\omega_f = 0.42 \;rad/s[/tex]

b. To calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.

Before the person's walk:

[tex]K.E_R =\frac{1}{2} I_i\omega_i^2\\\\K.E_R =\frac{1}{2} \times 820 \times 0.85^2\\\\K.E_R =410\times 0.7225\\\\K.E_R = 296.23 \;Joules[/tex]

After the person's walk:

[tex]K.E_R =\frac{1}{2} I_f\omega_f^2 + \frac{1}{2} I_f\omega_f^2\\\\K.E_R =\frac{1}{2} \times 820 \times 0.42^2 + \frac{1}{2} \times 75(3.3)^2 \times 0.42^2\\\\K.E_R =410\times 0.1716 + 408.375\times 0.1716\\\\K.E_R = 70.36+70.08\\\\K.E_R = 140.44 \;Joules[/tex]

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