my phy proff forgot to post the class lectures online (we dont do lectures in class... long story) so i dont kno how to do any of the problems given on his online quiz... please help
A bobsled at 162.1 km/hr turns around a curve with a 48.12 m radius of curvature on a banked ice track. What is the banking angle if banking alone (friction neglected) is to hold the sled on its track
answer is 76.89348616804996 ... but i need to understand how to get it (equation and methods)
2007-02-20 00:59:13 · 3 answers · asked by Paul S 2 in Science & Mathematics ➔ Physics
banking angle=tan^-1(v^2/rg)
put v=162.1km/hr
r=48.12 m
g=9.8m/s^2
therefore,
banking angle=76.89348616804996 degree
2007-02-20 01:04:59 · answer #1 · answered by Anonymous · 0⤊ 0⤋
There are 2 forces, gravity and widely used rigidity (N). you like them to characteristic to offer a horizontal rigidity that effects in the dazzling centripetal acceleration. The vertical element of the conventional rigidity lots experience gravity, so N cos theta = mg The horizontal element could produce the centripetal acceleration, so N sin theta = mv^2/r Divide the 2nd equation with the help of the 1st to get tan theta = v^2/gr tan theta = 24.5^2/(9.8*one hundred ten) = 0.fifty six theta = 29 stages
2016-09-29 09:03:35 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Use the formula
tan @=v^2/gr
where v=angular speed
g=gravitational acceleration
r=radius
Good luck!!anyway u r from which country?
2007-02-20 01:27:59 · answer #3 · answered by yan 2 · 0⤊ 0⤋