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?

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

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