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Grade: 9
        
A roller of mass 300 kg and of radius 50 cm s lying on  horizontal floor is resting against a step of height 20 cm . the minimum force to be applied on the roller passing through its centre to turn the roller on the step?
2 years ago

Answers : (2)

Roshan Yarlagadda
34 Points
							
Mgh=1/2mv2+1/2Iw2
mgh=1/2mv2+1/2mr2v2/r2
mgh=mv2
v=root gh
v=1.414 m/s
again
W=F.v
1/2mv2+1/2mr2v2/r2=F.v
mv2=F.v
F=m.v
F=300*1.414
F=424.2 N
2 years ago
bharat
29 Points
							
i think that for a rolling on a rough inclined plane might help u ,since if these are the steps of ladder then we will have to make equations of motion of rotating body with constant acceleration and also force and mg equations.
if it is inclined then :
acceleration at c.m.=(gsin\Theta/1+I/mr^2)
where \Theta is the angle of the inclined plane  and I is moment of inertia .
please let me know if that step is of ladder or is of incline plane, 
i will make sure that u get the answer of the question afterwards
 
 
2 years ago
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  • NCERT Solutions
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  • Previous Year Exam Questions


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