Mechanics> A string breaks on suspending a 50KG mass...

A string breaks on suspending a 50KG mass.One Kg mass is tied to one end of the 10m long string and revolved in a
horizontal plane. find the miximum number of rotations per minute so that the string may not break.
1)67
2)57
3)47
4)97

Saurabh , 8 Years ago

Grade 12

2 Answers

Tejus

we known critical mass is 500 N

mw²r= 500

w²=50

we want to have no. of rotations per minute I.e.

no. of 2py radians per 60 sec

let no. of rotation = n

{ n× 2 py/60 }²= 50 on calculating n comes to 67.52

=67 rotations approximately

I don't known whether I did wrong or right as I did it in my first attempt

only the last line was tricky rest easy one

thank you

Approved

Last Activity: 8 Years ago

Arun

Length r= 10 m

Mass m= 1 kg

Centripetal force F = 50 kg wt = 50* 9.8 N

Let n be the maximum number of revolutions made per

second without breaking string.

Then F = m*r*w²= m r (2 $\pi$ n)²

50 * 9.8 = 1*10 (44/7 * n)²

490 = 395.10 n²

n = 1.11 revolution per second

Hence revolution per minute = 1.11* 60 = 66.66 = 67

Approved

Last Activity: 8 Years ago

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Mechanics> A string breaks on suspending a 50KG mass...

A string breaks on suspending a 50KG mass.One Kg mass is tied to one end of the 10m long string and revolved in ahorizontal plane. find the miximum number of rotations per minute so that the string may not break.1)672)573)474)97

Saurabh , 8 Years ago

Tejus

Arun

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