Physics, 01.08.2019 02:20 Shiey0409

Adisk of mass m and radius r is initially held at rest just above a larger disk of mass m and radius r that is rotating at angular speed ωi. what is the final angular speed of the disks after the top one is dropped onto the bottom one and they stop slipping on each other?

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Answer from: cmfield8810

$\frac{MR^{2}\omega _{i}}{(MR^{2}+mr^{2})}=\omega _{f}$

Explanation:

Mass of small disc = m

radius of small disc = r

Mass of large disc = M

Radius of large disc = R

Initial angular speed = ωi

Let the final angular speed is ωf

As no external torque of applied on the system so the angular system remains constant.

I1 ωi = I2 ωf

MR^2 ωi = (MR^2 + mr^2) ωf

$MR^{2}\omega _{i}=(MR^{2}+mr^{2})\omega _{f}$

$\frac{MR^{2}\omega _{i}}{(MR^{2}+mr^{2})}=\omega _{f}$

Answer from: Quest

Electoral current force mechanical advantage if the wheel and axle is 2 n and the output force is 20 n

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

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Adisk of mass m and radius r is initially held at rest just above a larger disk of mass m and radius...

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