Question

A flywheel in the shape of uniform disc has a mass 40kg and is of radius 0.5m. It is revolving around

its own axis at the rate of 360rpm. What torque is needed to bring it to rest in 10 s? If the torque is on

account of a force applied tangentially on the rim of flywheel, what is magnitude of force.



please give the correct answer otherwise your answer will be reported❤❤❤❤❤​

Answer 1

Answer:

Torque = 12 π Nm

Force = 24 π N

Explanation:

Given:

• Mass of the disc = 40 kg
• Radius of the disc = 0.5 m
• Initial angular velocity = 360 rpm  
• Time taken = 10 s

To Find:

• Torque required to bring it to rest in 10 s

Solution:

Let ω₀ be the initial angular velocity and ω be the final velocity.

ω₀ = 360 rpm = 360 × π/30 = 12 π rad/s

Now finding the angular acceleration α,

By the first equation of kinematics we know that,

ω = ω₀ + αt

Substituting the data we get,

0 = 12 π + α × 10

α = -12 π/10

α = -1.2 π rad/s²

Finding the moment of inertia I,

I = MR²

I = 40 × (0.5)²

I = 10 kg m²

Now the torque produces is given by,

τ = I α

τ = 10 × 1.2 π = 12 π Nm

Hence the torque produced to bring the flywheel to rest is 12 π Nm.

Given that force is applied tangentially,

τ = R F

F = τ/R

F = 12 π/0.5

F = 24 π N

Hence the magnitude of the force is 24 π N.

Answer 2

Answer:

Given :-

• A flywheel in the shape of uniform disc has a mass of 40 kg and it's radius is 0.5 m. It is revolving around its own axis at the rest of 360 rpm. The torque is needed to bring it to rest in 10 seconds.

To Find :-

• What is the magnitude of force.

Formula Used :-

➦ ω = ω₀ + αt

➦ I = MR²

➦ τ = I α

➦ F = τ/R

Solution :-

First we have to find the initial angular velocity,

Given :

• ω₀ = 360 rpm

Then,

↦ ω₀ = 360 rpm

↦ ω₀ = 360 × π/30 rps

➠ ω₀ = 12 π rad/s

Again, we have to find the angular acceleration,

Given :

• ω = 0
• ω₀ = 12 π
• t = 10 seconds

According to the question by using the formula we get,

↦ 0 = 12 π + α(10)

↦ 0 = 12 π + 10α

↦ α = - 12 π/10

➠ α = - 1.2 π rad/s²

Again, we have to find the moment of inertia,

Given :

• Mass = 40 kg
• Radius = 0.5 m

According to the question by using the formula we get,

↦ I = 40 × (0.5)²

↦ I = 40 × 0.25

↦ I = 40 × 25/100

↦ I = 1000/100

➠ I = 10 kg m²

Again, we have to find the torque,

Given :

• l = 10 kg m²
• α = 1.2 π rad/s²

According to the question by using the formula we get,

↦ τ = 10 × 1.2

↦ τ = 10 × 12/10

↦ τ = 120/10

➠ τ = 12 Nm

Now, we have to find the magnitude of the force,

Given :

• τ = 12 Nm
• Radius = 0.5 m

According to the question by using the formula we get,

⇒ F = 12/0.5

⇒ F = 12 × 10/5

⇒ F = 120/5

➤ F = 24 N

∴ The magnitude of force is 24 N .