Answer to Question #118531 in Mechanics | Relativity for Jeshneel Kumar

Question #118531
A flywheel has a moment of inertia of 1.6 × 10−3 kg.m2
. When a constant torque is
applied, it reaches an angular velocity of 1200 rev/min in 15 s. Assuming it started from
rest, find:
i) the angular acceleration
ii) the unbalanced torque applied
iii) the angle turned through 15 s
iv) the work done on the flywheel by the torque
1
Expert's answer
2020-05-27T10:50:44-0400

Solution.

"I=1.6\\sdot10^{-3}kg.m^2;"

"n=1200rev\/min=20rev\/s;"

"t=15s;"

w_0=0;

i)"\\alpha=\\dfrac{\\Delta w}{t}; \n\\Delta w=w-w_0;"

"w=2\\pi n;"

"w=2\\sdot3.14rad\\sdot20rew\/s=125.6rad\/s" ;

"\\alpha=\\dfrac{125.6rad\/s}{15s}=8.37rad\/s^2;"

ii) "\\alpha=\\dfrac{T}{I};\\implies T=\\alpha I;"

"T=8.37rad\/s^2\\sdot1.6\\sdot10^{-3} kg.m^2=0.013kg.m^2\/s^2;"

iii)"\\theta=\\theta_0+w_0t+\\dfrac{\\alpha t^2}{2};"

"\\theta=\\dfrac{8.37rad\/s^2\\sdot(15s)^2}{2}=941.6rad;"

iv)"W=\\dfrac{1}{2}Iw^2-\\dfrac{1}{2}Iw_0^2;"

"W=\\dfrac{1}{2}\\sdot1.6\\sdot10^{-3}kg.m^2\\sdot(125.6rad\/s)^2="

"=12.62J;"

Answer: i)"\\alpha=8.37rad\/s^2;"

ii)"T=0.013kg.m^2\/s^2;"

iii)"\\theta=941.6rad;"

iv)"W=12.62J."



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