Asked by Amber
Consider a wheel (solid disk) of radius
0.748 m, mass 16 kg and moment of iner-
tia 1/2 MR^2 . The wheel rolls without slipping in a straight line in an uphill direction 27 above the horizontal. The wheel starts at
angular speed 20.5882 rad/s but the rotation slows down as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill.How far does the wheel roll in the uphill
direction before it stops? The acceleration of gravity is 9.8 m/s^2 .
Answer in units of m
0.748 m, mass 16 kg and moment of iner-
tia 1/2 MR^2 . The wheel rolls without slipping in a straight line in an uphill direction 27 above the horizontal. The wheel starts at
angular speed 20.5882 rad/s but the rotation slows down as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill.How far does the wheel roll in the uphill
direction before it stops? The acceleration of gravity is 9.8 m/s^2 .
Answer in units of m
Answers
Answered by
Elena
KE = PE,
KE = mv^2/2 + Iω^2/2 =m• (ω•R)^2/2 = m•R^2• ω^2/4 = 3• m•R^2• ω^2/4,
PE = m•g•h = m•g•s•sinα,
3• m•R^2• ω^2/4 = m•g•s•sinα,
s =3•R^2• ω^2/4•g•sinα.
KE = mv^2/2 + Iω^2/2 =m• (ω•R)^2/2 = m•R^2• ω^2/4 = 3• m•R^2• ω^2/4,
PE = m•g•h = m•g•s•sinα,
3• m•R^2• ω^2/4 = m•g•s•sinα,
s =3•R^2• ω^2/4•g•sinα.
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