contestada

The 2 kg 30 cm diameter disk is spinning at 300 rpm. How much friction force must the brake appply to the rim to bring the disk to a halt?

Respuesta :

Khoso123

Answer:

Force of Friction= 1.57 N

Explanation:

Given Data

Mass=2 kg

Diameter=30 cm

To Find

Friction Force=?

Solution

First We need to find angular deceleration rate

a = (300×2π/60 rad/s)/3.0 s

a= (10/3)×π = 10.47 rad/s²

The required torque T is given by

T = I×a

where

I is the moment of inertia of the disc, (1/2) M×R² (M is mass and R is radius)

a is angular deceleration

Therefore

T= (1/2) M×R² ×10.47

first we need to find radius in meter so

R=diameter/(2×100)

R=(30 cm)/(2×100) m

R=0.15 m

T=(1/2)(2 kg)×(0.15 m)²(10.47 rad/s² )

T=0.2355 N.m

Once you have determined torque T, use the below formula to find friction

Force of Friction= T/R

Force of Friction= 0.2355 N.m/0.15 m

Force of Friction= 1.57 N

onyebuchinnaji

The frictional force applied to the rim to bring the disk to halt is 592.33 N.

The given parameters;

  • mass of the disk, m = 2 kg
  • radius of the disk, r = 30 cm = 0.3 m
  • angular speed, ω = 300 rpm

The angular speed is calculated as follows;

[tex]\omega = 300 \ \frac{rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \\\\\omega = 31.42 \ rad/s[/tex]

The frictional force applied to the rim to bring the disk to halt is calculated as follows;

[tex]F = m \omega ^2 r\\\\F = (2) \times (31.42)^2 \times (0.3)\\\\F = 592.33 \ N[/tex]

Thus, the frictional force applied to the rim to bring the disk to halt is 592.33 N.

Learn more here:https://brainly.com/question/13903806

Otras preguntas