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A uniform disk with a rotational inertia of 2.0 kg*m^2 and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicular to the disk faces. A uniform force of 4.0 N is applied tangentially to the rim of the disk. What is the angular velocity of the disk after 6.0 s?

A) 4.8 rad/sB) 1.8 rad/sC) 18 rad/sD) 3.6 rad/sE) 12 rad/s

Respuesta :

Answer:

B)ωf = 1.8 rad/s

Explanation:

Given that

I= 2 kg.m²

d= 30 cm ,r= 15 cm

F= 4 N

t= 6 s

The torque produce by force F about point center of disc

T= F .r

T= 4 x 0.15 = 0.6 N.m

Lets take angular acceleration of the disc is α rad/s²

T= I α

0.6 = α x 2

α= 0.3 rad/s²

We know that

ωf = ωi + α t

ω =Angular speed of the disc

Initial angular speed ωi = 0 rad/s

ωf = 0 + 0.3 x 6

ωf = 1.8 rad/s

Therefore the answer is B.

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