Respuesta :
Answer:
Given that
r= 0.2 m
m= 40 kg
F= 30 N
We know that mass moment of inertia of disk given as
I = m r²/2
I= 40 x 0.2²/2
I= 0.8 kg.m²
The torque produce by force
T = F.r
T= 30 x 0.2 = 6 N.m
We also know that
T= I α
α = angular acceleration of disk
by putting the values
6 = 0.8 x α
α = 7.5 rad/s²
Given that initial speed is zero.
Lets speed is ω after 0.2 rev
ω ² = ωo+ 2 α θ
We know that
1 rev = 2π
0.2 rev= 0.4π
θ = 0.4π
ω ² = 0+ 2 x 7.5 x 0.4 x π
ω = 4.34 rad/s
So the tangential velocity v
v= ω r
v= 4.34 x 0.2 = 0.86 m/s
v= 0.86 m/s
Tangential acceleration (at)= α .r= 7.5 x 0.2 =1.5 m/s²
Radial acceleration(ar) = ω² r = 4.34² x 0.2 = 3.76 m/s²
resultant acceleration ,a
[tex]a=\sqrt{at^2+ar^2}[/tex]
[tex]a=\sqrt{1.5^2+3.76^2}[/tex]
a = 4.05 m/s²
The magnitude of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.200 revolutions is 4.05 m/s².
What is acceleration?
Acceleration is the rate of change of velocity with respect to time.
As it is given that the radius(r) of the disk is 0.2 m, while the mass of the disk is 40 kg, and the constant force applied is 30 N.
Now since we know that mass moment of inertia of disk is given as
[tex]I = \dfrac{m r^2}{2}\\\\\text{Substituting the values}\\\\I = \dfrac{40 \times 0.2^2}{2}\\\\I = 0.8\ kg\cdot m^2[/tex]
Thus, the mass moment of inertia of the disk is 0.8 kg·m².
The torque produced by force on the disk can be written as,
[tex]T = F\cdot r = I \times \alpha \\\\30 \times 0.2 = 0.8 \times \alpha \\\\ \alpha = 7.5\rm\ rad/sec^2[/tex]
Thus, the acceleration of the disk can be written as 7.5 rad/s².
The circular distance covered by the disk can be written as,
[tex]\rm 1\ rev = 2\pi \\\\0.2\ rev= 0.4\pi \\\\\theta = 0.4\pi[/tex]
As it is given that the initial speed is zero. therefore, according to the second equation of motion for the circular motion can be as
[tex]\omega^2 = \omega_o+ (2\alpha\ \theta) \\\\\omega^2 = 0+ (2 \times 7.5 \times 0.4 \times \pi)\\\\\omega = 4.34\ rad/s[/tex]
Now the tangential velocity v can be written as,
[tex]v=\omega \times r\\\\v= 4.34 \times 0.2 \\\\v= 0.86 m/s[/tex]
Further, the components of the acceleration can be written as,
Tangential acceleration
[tex]a_t= \alpha \times r\\\\a_t= 7.5 x 0.2 \\\\a_t=1.5\rm\ m/s^2[/tex]
Radial acceleration
[tex]a_r = \omega^2 r \\\\a_r= 4.34^2 \times 0.2 \\\\a_r= 3.76\rm\ m/s^2[/tex]
Resultant acceleration
[tex]a = \sqrt{a_t^2+a_r^2}\\\\a = \sqrt{1.5^2+3.76^2}\\\\a = 4.05\rm\ m/s^2[/tex]
Hence, the magnitude of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.200 revolutions is 4.05 m/s².
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