Question

A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius

Answer 1

Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

          KE = 0.23 J

given parameters

• Disk radius R = 15 cm = 0.15 m
• Cylinder radius r = 1.5 cm = 0.0015 m
• Disk mass M = 20 kg
• Time t = 1.2 s
• Force F = 12 N

to find

• Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

         KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

         I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk          I_{disk} = ½ M R²

cylinder   I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

         I_{total} = ½ (M R² + m r²) = ½ M R² ( $1 + \frac{m}{M} \ (\frac{r}{R})^2$ )

         I_{total} = ½ M R² ( $1+ \frac{m}{20} (\frac{0.015}{0.15} )^2$ ) = $\frac{1}{2}$ M R² (1 + 0.005 m)

As the shaft mass  is much lighter than the disk mass , the last term is very small, which is why we despise it.

         I_{total} = ½ M R²

2nd part. Let's use Newton's second law for endowment motion

        τ = I α

        α = $\frac{\tau }{I_{total}}$l

        τ = F R

        α = $\frac{F \ R}{I_{total}}$

With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)

        w = w₀ + α  t

where w is the angular velocity, alpha is the angular acceleration and t is the time

        w = 0 + $\frac{\tau }{I_{total}} \ t$

we substitute in the kinetic energy equation

        KE = ½ I_{total}  ( $\frac{ \tau }{I_{total}} \ t$ )²

        KE = ½ $\frac{ \tau^2 }{I_{total}} \ t^2$

let's substitute

        KE = $\frac{F^2 \ R^4}{M \ R^2 } \ t^2$

        KE = F² R² t² / M

let's calculate

        KE = 12² 0.15² 1.2² / 20

        KE = 0.23 J

With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is

       KE = 0.23 j

learn more about rotational kinetic energy here:

brainly.com/question/20261989