Question

The radius of gyration of a solid disc wheel of uniform thickness is at r/sqrt2from the center, where ris the radius of the wheel. Calculate the change in kinetic energy, in kilo joules, in a solid disc flywheel of 1400 mm diameter and 1250Kg mass when its speed changes from 8 pie to 10 pie rad/sec.(Answer In KJ)

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Answer 1

Answer:

• The Kinetic Energy is 54.02 KJ

Explanation:

Given, Radius of gyration (K) = r/√2

Diameter (d) = 1400mm = 1.4 m

Mass of solid disc (m) = 1250 Kg

Initial angular veloc. (ω₁) = 8π rad/sec

Final angular veloc. (ω₂) = 10π rad/sec

From the formula of we know that,

$\longrightarrow\sf K.E. = \dfrac{1}{2}\;.\;I\;.\;\omega^{2}$

Substituting the values,

$\longrightarrow\sf K.E = \dfrac{1}{2}\times \Bigg(M\:K^{2}\Bigg)\times \bigg(\omega_{2}^{2}-\omega_{1}^{2}\bigg)\\\\\\\\\longrightarrow\sf K.E = \dfrac{1}{2}\times \Bigg(1250\times \bigg(\dfrac{r}{\sqrt{2}}\bigg)^{2}\Bigg)\times \bigg((10\pi)^{2}-(8\pi)^{2}\bigg)\\\\\\\\\longrightarrow\sf K.E = \dfrac{1}{2}\times \Bigg(1250\times \dfrac{r^{2}}{2}\Bigg)\times \bigg(100\;\pi^{2}-64\;\pi^{2}\bigg)$

Diameter = 1.4 m so, radius is 0.7 m.

$\longrightarrow\sf K.E = \dfrac{1}{2}\times \Bigg(625\times (0.7)^{2}\Bigg)\times \bigg(36\;\pi^{2}\bigg)\\\\\\\\\longrightarrow\sf K.E = \Bigg(625\times 0.49\Bigg)\times \bigg(18\;\pi^{2}\bigg)\\\\\\\\\longrightarrow\sf K.E = 306.25 \times 18\;\pi^{2}\\\\\\\\\longrightarrow\sf K.E = 54022.5 = 54.022 \times 10^{3}\\\\\\\\\longrightarrow\sf K.E = 54.02 \; KJ$

∴  The Kinetic Energy of solid disc is 54.02 KJ.

Answer 2

Question:-

The radius of gyration of a solid disc wheel of uniform thickness is at r/√2 from the center, where ris the radius of the wheel. Calculate the change in kinetic energy, in kilo joules, in a solid disc flywheel of 1400 mm diameter and 1250Kg mass when its speed changes from 8π to 10π rad/sec.(Answer In KJ)

To Find:-

• changes in the kinetic energy.

Given:-

• radius of gyration (k) = r/√2.
• Diameter (d) = 1400mm.
• Mass of solid disc (m) = 1250Kg.
• Initial angular velocity (ω₁) = 8π rad/sec.
• Final angular veloc. (ω₂) = 10π rad/sec.

Formula:-

From the Formula we know that,

$\sf{=> K.E. = \frac{1}{2} .I .ω {}^{2} }$

Solution:-

$\sf{radius \: of \: gyration = \sqrt{ \frac{I}{A} } }$

$\sf{Area\: of \: disc \: = πr {}^{2} }$

$\sf{=> \frac{r}{ \sqrt{2} } = \sqrt{ \frac{I}{\pi r {}^{2} } } }$

$\sf{ = > \frac{r {}^{2} }{2} = \frac{</p><p>I}{\pi r {}^{2} } }$

$\sf{ = > I = \frac{\pi r {}^{4} }{2} }$

$\sf{\therefore Inertia \: of \: the \: disc \: is \: = \frac{\pi r {}^{4} }{2} }.$

$\sf{kinetic \: energy \: of \: a \: disc \: is \: given \: by} \\ \sf{ \frac{1}{2} }mr {}^{2} ω {}^{2} + \frac{1}{2} \: I \: ω {}^{2}$

$\sf{=> change \: in \: kinetic \: energy \: is}$

$\sf{= \frac{1}{2} mr^2 (ω₂ {}^{2} \: – \: {\sf{{ω_{1}}}} {}^{2} )+ \frac{1}{2} \: I } \: (ω₂ {}^{2} - {\sf{{ω_{1}}}} {}^{2} )$

$= [ \frac{1}{2} \times 1250 \times ( \frac{1400}{2 \times 1000} ) {}^{2} + \\ \frac{1}{2} \times\frac{\pi}{2} ( \frac{1400}{2 \times 1000} ) {}^{4} ]({\sf{{ω_{1}}}}{\sf{{ - ω_{2}}}})$

$= (306.25 + 0.1886)((10\pi) {}^{2} - \: \: \: \: \: \: \: \: \: \: ((8\pi) {}^{2} )$

$\sf{108,880 \: KJ(m/s) {}^{2} }$

$\sf{54440 \:Joules }$

$\sf{54.44 \: KJ \: also \: similarly \: written \: as \: } \\ \sf{ 54.02 KJ.}$

Answer:-

The changes in kinetic energy is 54.02 KJ.