Question

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

Answer 1

The angular velocity of the disk after 6.0 s, with a uniform force applied tangentially, is calculated to be 1.8 rad/s.

Step-by-step explanation:

To determine the angular velocity of the disk after a certain time, we can use the relationship between torque (τ), angular acceleration (α), and angular velocity (ω). The formula for torque is τ = Iα, where I is the moment of inertia and α is the angular acceleration. Once we have the angular acceleration, we can find the angular velocity using ω = ω0 + αt.

In this case, the torque caused by the uniform force applied tangentially to the rim of the disk can be calculated as τ = Fr, where F is the force and r is the radius of the disk. The radius is half of the diameter, so r = 0.15 m. Therefore, τ = 4.0 N × 0.15 m = 0.6 N·m.

Now, using the formula for torque, we solve for angular acceleration: 0.6 N·m = 2.0 kg·m2 × α, giving us α = 0.3 rad/s2. Finally, since we're looking for the angular velocity after 6.0 s from rest (ω0 = 0), we find the angular velocity ω = 0 + (0.3 rad/s2 × 6.0 s) = 1.8 rad/s.

Answer 2

B)ωf = 1.8 rad/s

Step-by-step 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.