Grade 12Modern Physics

# 7 A disk of diameter 15 cm is spinning at 3000 RPM. What is the acceleration of a point on the outside edge of the disk?

him sharma

12 Years agoGrade 12

1 Answer

Arun Kumar

12 Years ago

$\\\omega = 3000rpm=3000/60rad/sec \\=50*2\pi {rad \over sec}=100\pi {rad \over sec} \\=>\omega =100\pi {rad \over sec} \\a=a_c=\omega ^2*r \\=(100\pi)^2*0.15=151.98$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

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# 7 A disk of diameter 15 cm is spinning at 3000 RPM. What is the acceleration of a point on the outside edge of the disk?

him sharma

1 Answer

Arun Kumar

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