Question

on applying a torque,a flywheel acquired on angular speed of 50 revolution in 10.5 sec. if moment of inertia of flywheel is 5 kg m² then calculate the torque acts on it. ​

Answer 1

Answer:

Torque = 14.13 N-M

Explanation:

Given that,

a flywheel acquired on angular speed of 50 revolution in 10.5 sec

10.5 seconds = 50 revolutions

1 second = 50/10.5

= 4.76 revolution

so,

here,

Frequency of the revolution(f) = 4.76

so,

angular velocity = 2πf

= 9.52π rad/s

Here,

given the initial angular velocity(W•)

= 0 rad/s

and final angular velocity(W) =

9.52π tad/s

time taken(t) = 10.5 s

Now,

W = W• + αt

where,

α is the angular acceleration

putting the values,

9.52 = 0 + 10.5α

α = 9.52/10.5

α = 0.9 rad/s²

Now,

we have,

Torque = Iα

where,

I is the moment of inerta

that is

5 kg m²

putting the values,

Torque = 5 × 0.9

Torque = 4.5 π NM

π = 3.14

so,

torque = 4.5 × 3.14

Torque = 14.13 N-M

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Frequency of the revolution (f) = 50 rev/10.5sec.
• Moment of Inertia (I) = 5 Kgm².
• Time period (t) = 10.5 Seconds.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

As the frequency of revolution is given in the terms of 10.5 seconds, I.e.

$\large{\tt 10.5 \: Seconds = 50 revolutions}$

Then for 1 Second,

$\large{\tt 1 second = \dfrac{50}{10.5}}$

Now,

$\large{\tt 1 second = \dfrac{500}{105}}$

$\large{\tt 1 second = 4.76 }$

∴$\large{\boxed{\tt f = 4.76 \: rev/sec}}$

So, the Frequency of revolution (f) is 4.76 rev/seconds.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, To get Angular Velocity,

$\large{\boxed{\tt \omega = 2 \pi f}}$

Substituting the value,

$\large{\tt \omega = 2 \times 4.76 \times \pi}$

$\large{\boxed{\tt \omega = 9.52 \pi}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, Applying Angular Kinematic equation,

$\large{\boxed{\tt \omega = \omega_o + \alpha t}}$

• ω = Final Angular velocity
• ω₀= Initial Angular velocity
• α = Angular acceleration
• t = time taken.

Substituting the values,

(Initial angular velocity will be zero.)

$\large{\tt 9.52 \pi = 0 + \alpha \times 10.5}$

$\large{\tt 9.52 \pi = \alpha \times 10.5}$

$\large{\tt \alpha = \dfrac{9.52 \pi }{10.5}}$

$\large{\boxed{\tt \alpha = 0.90 \pi \: rad/sec^2}}$

So, The Angular acceleration is 0.90 π rad/sec².

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From The Moment of Inertia and Torque relation.

$\large{\boxed{\tt \tau = I \alpha}}$

Substituting the values,

$\large{\tt \tau = 5 \times 0.90 \pi}$

As π =3.14

$\large{\tt \tau = 5 \times 0.90 \times 3.14}$

$\large{\tt \tau = 4.5 \times 3.14}$

$\huge{\boxed{\boxed{\tt \tau = 14.13 \: Nm}}}$

So, The Torque that acts on the flywheel is 14.13 N-m.

$\rule{300}{1.5}$