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A string is wrapped around a uniform disk of mass M = 1.2 kg and radius R = 0.07 m. (Recall that the moment of inertia of a uniform disk is (1/2)MR2.) Attached to the disk are four low-mass rods of radius b = 0.11 m, each with a small mass m = 0.5 kg at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F = 29 N. At the instant when the center of the disk has moved a distance d = 0.039 m, a length w = 0.023 m of string has unwound off the disk.(a) At this instant, what is the speed of the center of the apparatus?v = ? m/s(b) At this instant, what is the angular speed of the apparatus?w1 = ? radians/s(c) You keep pulling with constant force 29 N for an additional 0.022 s. Now what is the angular speed of the apparatus?w2 = ? radians/s

Respuesta :

Explanation:

Given that,

Mass of disk = 1.2 kg

Radius = 0.07 m

Radius of rod = 0.11 m

Mass of small disk = 0.5 kg

Force = 29 N

Time t = 0.022 s

[tex]\theta=0.023\ m[/tex]

Distance d= 0.039 m

(I). We need to calculate the speed of the apparatus

Using work energy theorem

[tex]W=\Delta K.E[/tex]

[tex]Fd=\dfrac{1}{2}mv^2[/tex]

[tex]v=\sqrt{\dfrac{2Fd}{M+4m}}[/tex]

Where, m = total mass

v = velocity

F = force

d = distance

Put the value into the formula

[tex]v=\sqrt{\dfrac{2\times29\times0.039}{1.2+4\times0.5}}[/tex]

[tex]v=0.840\ m/s[/tex]

(b). We need to calculate the angular speed of the apparatus

Using formula of torque

[tex]\tau=I\alpha[/tex]

[tex]F\times=(\dfrac{1}{2}MR^2+4mb^2)\alpha[/tex]

[tex]29\times0.07=(\dfrac{1}{2}\times1.2\times0.07^2+4\times0.5\times0.11^2)\alpha[/tex]

[tex]\alpha=\dfrac{29\times0.07}{0.02714}[/tex]

[tex]\alpha=74.79\ rad/s^2[/tex]

We need to calculate the angular speed of the apparatus

Using equation of angular motion

[tex]\omega=\omega_{0}+\alpha t[/tex]

Put the value into the formula

[tex]\omega=0+74.79\times0.022[/tex]

[tex]\omega=1.645\ rad/s[/tex]

(c).  We need to calculate the angular speed of the apparatus

Using equation of angular motion

[tex]\omega_{0}^2=\omega^2+2\alpha t[/tex]

Put the value into the formula

[tex]\omega_{0}^2=1.645^2+2\times74.79\times0.022[/tex]

[tex]\omega=2.44\ rad/s[/tex]

Hence, This is required equation.

snehashish65

Answer:

The speed at center of apparatus is 0.840 m/s.

The angular speed of apparatus is 1.645 rad/s.

The angular speed for additional time is 3.290 rad/s.

Explanation:

Given data:

Mass of disk is, M = 1.2 kg.

Radius of disk is, R = 0.07 m.

Radius of rods is, b = 0.11 m.

Mass of rod is, m = 0.5 kg.

Magnitude of constant force is, F = 29 N.

Distance moved by disk is, d = 0.039 m.

Length of string is, L = 0.023 m.

Time interval is, t = 0.022 s

(a)

Speed at center of apparatus is obtained by work-energy theorem as,

[tex]W = \delta KE\\F \times d = \dfrac{1}{2}(M+4m)v^{2} \\29 \times 0.039 = \dfrac{1}{2}(1.2+(4 \times 0.5))v^{2} \\v = 0.840 \;\rm m/s[/tex]

Thus, the speed at center of apparatus is 0.840 m/s.

(b)

Torque at the instant is,

[tex]T = I \times \alpha[/tex]

Here, I is moment of inertia of system and [tex]\alpha[/tex] is the angular acceleration.

[tex]T = I \times \alpha\\F \times R = (\dfrac{1}{2}MR^{2}+4mb^{2} ) \times \alpha\\29 \times 0.07 = (\dfrac{1}{2} \times 1.2 \times 0.07^{2}+4 \times 0.5 \times 0.11^{2} ) \times \alpha\\\alpha = 74.79 \;\rm rad/s^{2}[/tex]

Then angular speed is,

[tex]\alpha = \dfrac{ \omega}{t} \\74.79 = \dfrac{ \omega}{0.022} \\\omega =1.645 \;\rm rad/s[/tex]

Thus, the angular speed of apparatus is 1.645 rad/s.

(c)

Angular speed for time t is,

[tex]\omega'=\omega+\alpha t\\\omega'=1.645+(74.79 \times 0.022)\\\omega'=3.290 \;\rm rad/s[/tex]

Thus, the angular speed for additional time is 3.290 rad/s.

For more details, refer the link:

https://brainly.com/question/15999056?referrer=searchResults

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