Question

A disc of mass 25kg and diameter 0.4m is rotating about its axis at 240 rev/min.Find the tangential force needed to stop it in 25sec.​

Answer 1

tangential force needed to stop is 5.02655N.

Answer 2

The Tangential force needed to stop the disc in 25 sec is 5.024 N

Step-by-step explanation:

Given Data

Mass of  the disc = 25 kg

Diameter of the disc = 0.4 m then  radius = 0.2 m

Speed $(w_o)$ = 240 rotation per minute

Time = 25 sec

Find the tangential force need to stop the disc

$w_0 = 240 \text rpm$

$w_0 = \frac{240}{60 \times 60} rps$

$w_0 = 0.067 rps$

$w_0 = 8 \pi rad s^{-1}$(radian per second)

From the relation, $w = w_0 + \alpha t$

Substitute the respective values in above equation.

0 = 8 π + α (25)

$\alpha = \frac{8 \pi}{25}$  

$\alpha = 0.32 \pi rad s^{-1}$

Moment of inertia of the disc I = mass × square of the radius

I = 25 × 0.2²

Moment of Inertia of the disc I = 1 kg m²

Torque required to stop the disc = Moment of inertia × α

Torque = 1 × 0.32 π

Torque = 1.0048 N-m

The tangential force need to stop the disc can be calculated by divide the Torque by radius of the disc.

$Tangential Force =\frac { \text { Torque }}{\text { Radius of the disc }}$

$\text { Tangential Force }=\frac{ 1.0048 }{0.2}$

Tangential force = 5.024 N

The tangential  force needed to stop the 25 kg disc rotates at 240 revolutions per minute is 5.024 Newton.

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