Question

g rll The 21-kg roll of paper has a radius of gyration ka = 90 mm about an axis passing through point A. It is pin supported

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Answer #1

Linear displacement;

s=vot + at

=1=0 + 5a(3)

> a=0.222 m/s

Angular acceleration;

a=_ 0.222 14- = 1.778 rad/s 0.125

By equilibrium of forces;

> F= 0

= N - TAB = 0

N=0.3846TAB

and

0 = 1

==TAB – 0.31. - 21(9.81) - F = 0

= 0.9231 TAB -0.3(0.3846TAB) – 21(9.81) - F = 0

F=0.8077TAB – 206.01

also

ΣΜΑ = Ια

-0.31.(0.125) + F(0.125) = 210.090)(1.778)

= -0.3(0.3846T AB)(0.125)+(0.8077TAB-206.01) (0.125) = 21(0.090)-(1.778)

+TAB = 301.059 N

Use equation (1);

F = 37.16 N

...(Answer)

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