Physics, asked by Anonymous, 10 months ago

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physics lovers??

a point on the rim of a flywheel and has a peripheral speed of 10 metre per second at an instant when it is decreasing at a rate of 60 metre per second square if the magnitude of the total acceleration of the point at this instant is hundred metre per second square the radius of the wheel is...??

Answers

Answered by Anonymous

Answer:-

Radius of wheel is = 1.25 m

Step - by - step explanation:-

Given :-

Total acceleration (A)= 100 m/s^2

speed (v)= 10 m/s

Tangential acceleration (A')= 60 m/s

Solution:-

Let centripetal acceleration= C

We know that,

$\bf{total \: acceleration(A) = \sqrt{ {c}^{2} + {A'}^{2} } } \\ \\ \bf{100 = \sqrt{ {c}^{2} + 3600 } } \\ \\ \bf{ {c}^{2} = 6400} \\ \\ \bf{ \: c = 80 \: \frac{m}{ {s}^{2} } }$

Now ,

We know that ,

$\bf{c = \frac{ {v}^{2} }{R} } \\ \\ \bf{ \: R = \frac{ {v}^{2} }{c} } \\ \\ \bf{ R = \frac{10 \times 10}{80} } \\ \\ \bf{ R = \frac{10}{8} } \\ \\ \bf{R = 1.25 \: m}$

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