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.
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.
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