Asked by jake
A solid sphere rolls without slipping from rest down a plane at 4.00m long inclined at an angle of 42 degrees with respect to the horizontal as shown in the figure below. Determine the speed at the bottom of the incline.
Answers
Answered by
Elena
PE=KE=KE(trans) +KE(rot)
PE= mgh=mgLsinα
KE= mv²/2 + Iω²/2=
= 2 mv² /2 + 2mr²v²/5•2•r²=
=0.5 mv² + 0.2 mv² =0.7 mv²
mgLsinα= 0.7 mv²
v =sqrt{gLsinα/0.7} =
= sqrt{9.8•4•sin42⁰/0.7} =6.12 m/s
PE= mgh=mgLsinα
KE= mv²/2 + Iω²/2=
= 2 mv² /2 + 2mr²v²/5•2•r²=
=0.5 mv² + 0.2 mv² =0.7 mv²
mgLsinα= 0.7 mv²
v =sqrt{gLsinα/0.7} =
= sqrt{9.8•4•sin42⁰/0.7} =6.12 m/s
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