Answer:
Given information is summarized as below:
[tex]\omega _{o}=0.270rev/sec=1.69 rad/sec\\\alpha =0.895rev/sec=5.62rad/sec\\d=0.790m[/tex]
1)Using first equation of kinematics for angular motion we have
[tex]\omega _{f}=\omega _{o}+\alpha t\\\\\therefore \omega _{f}=1.69+5.62\times 0.194\\\omega _{f}=2.780rad/sec[/tex]
2) Using second equation of kinematics for angular motion we have
[tex]\theta =\omega _{o}t+\frac{1}{2}\alpha t^{2}\\\\\theta =1.69\times 0.194+05\times 5.62\times0.194^{2}\\\\\theta = 0.4336rad[/tex]
3) The tangential speed is given as
[tex]v_{t}=\omega _{f}\times r\\\\v_{t}=2.78\times\frac{0.790}{2}=1.098m/s[/tex]
4)
The resultant acceleration is given by
[tex]a_{res}=\sqrt{(a_{tangential})^{2}+(a_{radial})^{2}}\\\\a_{res}=\sqrt{(\alpha r)^{2}+(\omega _{f}^{2}r)^{2}}\\\\a_{res}=3.77m/s^{2}[/tex]
The centripetal force is the force that keeps an object moving along a circular path.
Using the formula;
ωf = ωi - αt
Where;
ωf = final angular velocity
ωi = Initial angular velocity
α = angular acceleration
t = time taken
Substituting values;
ωf = 0.270 rev/s + (0.895 rev/s^2 × 0.194s)
ωf = 0.443 rev/s
Given that;
ω = θ/t
0.443 rev/s = θ/ 0.194s
θ = 0.443 rev/s × 0.194s
θ = 0.086 rev
The tangential speed is obtained from
v = rω
Since 1 rev/s = 2π rad/s
0.443 rev/s = 0.443 rev/s × 2π rad/s/1 rev/s
= 2.78 rad/s
v = (0.790 m/2) × 2.78 rad/s
v = 1.098 m/s
ac = v^2/r
ac= ( 1.098 m/s)^2/0.395
ac = 3.05 m/s^2
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