Question

A disc of mass 5 kg and radius 50cm rolls on the ground at the rate of 10m/s calculate k. E

Answer 1

$mass \: = m = 5kg$

$radius \: = r = 50cm = \frac{50}{100} = 0.5m$

$velocity \: = v = 10ms {}^{ - 1}$

$moment \: of \: inertia \: of \: disk \\ = m.l = \frac{1}{2} mr {}^{2} \\ \\ = \frac{1}{2} \times 5 \times 0.5 \times 0.5 \\ \\ = 0.625kgm {}^{2} \\ \\ total \: k.e \: = \frac{1}{2} lw {}^{2} + \frac{1}{2} mv {}^{2} \\ \\ = \frac{1}{2} l( \frac{v}{r {}} ) {}^{2} + \frac{1}{2} mv {}^{2} \\ \\ = \frac{1}{2} \times 0.625 \times ( \frac{10}{0.5} ) {}^{2} + \frac{1}{2} \times 5 \times 10 {}^{2} \\ \\ = 125 + 250 \\ \\ = 375 \: j \\ \\ \\ \\ threfore \: \: total \: k.e \: is \: 375 \: j$

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Answer 2

Total Kinetic energy is 375 J.

Step-by-step explanation:

Given : A disc of mass 5 kg and radius 50 cm rolls on the ground at the rate of 10 m/s.

To find : Calculate K. E ?

Solution :

Mass is m=5 kg

Radius is r=50 cm=0.5 m

Velocity is v=10 m/s

Moment of Inertia of Disk is given by,

$M=\frac{1}{2}m r^2$

$M=\frac{1}{2}\times 5\times (0.5)^2$

$M=0.625\ kgm^2$

Total Kinetic energy is given by,

$KE=\frac{1}{2}(M(\frac{v}{r})^2+mv^2)$

$KE=\frac{1}{2}(0.625(\frac{10}{0.5})^2+(5)(10)^2)$

$KE=\frac{1}{2}(250+500)$

$KE=\frac{1}{2}(750)$

$KE=375\ J$

Therefore, Total Kinetic energy is 375 J.

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