A steady horizontal force F if magnitude 21N is apllied at the axle of a solid disk. the disk has mass 2.0kg and diameter of 0.10m. What is the linear speed of the center of the disk after it has moved 12m?
I wanted to use torque = torque
F*d = I*a
I= 1/2m*r^2
where a = v^2/r
F*d = (1/2 *m*r^2)(v^2/r)
still didn't get the answer its supposed to be B. 13 m/s
Well, if we're talking about a solid disk, we better be on solid ground here!
To find the linear speed of the center of the disk, we can use good old Newton's second law – F = ma.
First, let's find the acceleration of the disk. The force applied at the axle is the same as the torque applied, given by τ = Fr, where r is the radius of the disk. In this case, r is half the diameter, so r = 0.10m/2 = 0.05m.
Now, the torque applied τ = Iα, where I is the moment of inertia of the disk and α is the angular acceleration. For a solid disk, I = (1/2) * m * r^2, where m is the mass of the disk and r is the radius.
With a little algebraic acrobatics, we get α = (2 * τ) / (m * r^2).
Given that the disk has mass 2.0 kg and the applied force F = 21 N, we can calculate the torque τ = F * r = 21 N * 0.05 m = 1.05 Nm.
Plugging in the values, we get α = (2 * 1.05 Nm) / (2.0 kg * (0.05 m)^2) = 42 rad/s^2.
Now, we can use one of the kinematic equations to find the final linear speed of the center of the disk after it has moved 12m. We'll use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Since the disk starts from rest, u = 0, and the equation simplifies to v = sqrt(2 * a * s). Plugging in the values, we get v = sqrt(2 * 42 rad/s^2 * 12 m) ≈ 14.69 m/s.
So, after all this fun, the linear speed of the center of the disk is approximately 14.69 m/s. "Disk" out!
To find the linear speed of the center of the disk, we can use the principle of work and energy. The work done by the force F is equal to the change in kinetic energy of the disk.
The work done by a constant force is given by the formula:
Work = Force * Distance * cos(theta)
In this case, the force is applied horizontally, so the angle theta between the force and the direction of motion is 0 degrees. Therefore, cos(theta) is equal to 1.
The work done by the force F is:
Work = 21N * 12m * cos(0) = 252 J
The work done is equal to the change in kinetic energy:
Work = Change in Kinetic Energy
The kinetic energy of an object is given by the formula:
Kinetic Energy = 1/2 * Mass * Velocity^2
The initial kinetic energy of the disk is zero since it starts from rest. Therefore, the change in kinetic energy is equal to the final kinetic energy:
252 J = 1/2 * 2.0kg * Velocity^2
Solving for velocity:
Velocity^2 = (2 * 252 J) / (2.0kg)
Velocity^2 = 252 J / 2.0kg
Velocity^2 = 126 m^2/s^2
Velocity = √(126 m^2/s^2)
Velocity ≈ 11.22 m/s
So, the linear speed of the center of the disk after it has moved 12m is approximately 11.22 m/s.
Nick S. ? i got 15.87 m/s.... apparently the right answer is 9 m/s tho .
i assumed there's no torque.
v^2 = 2 ax. a = F / m.
B.K is an idiot
dont make of mrs keller shez da coolest, im tellin on u neil foglio and nick sun
peace