Search l 6:50 PM C @ 1 * 27% capa8.phy.ohio.edu Rolling down a Roof Due in 5 hou

Question

Search l 6:50 PM C @ 1 * 27% capa8.phy.ohio.edu Rolling down a Roof Due in 5 hours, 9 minutes A solid cylinder of radius 34.0 cm and mass 10.0 kg starts from rest and rolls without slipping a distance of 3.00 m down a roof that is inclined at 22°. (See the figure.) What is the angular speed of the cylinder about its center as it leaves the house roof? Submit Answer Tries 0/10 The roof's edge is 12.0 m high. How far horizontally from the edge of the roof does the cylinder hit the level ground? Submit AnswerTries 0/10 Post Discussion Send Feedback

Explanation / Answer

as the cylinder rolls on the roof

initial potential energy of cylinder PE = m*g*d*sintheta

at the edge of roof the cylinder has KE

KE = KEtrans + KErot

KE = (1/2)*m*v^2 + (1/2)*I*w^2

I = moment of inertia = (1/2)*m*R^2

w = angular speed = v/R

KE = (1/2)*m*v^2 + (1/2)*(1/2)*m*R^2*(v/R)^2

KE = (1/2)*m*v^2 + (1/4)*m*v^2

KE = (3/4)*m*v^2

from energy conservation PE = KE

(3/4)*m*v^2 = m*g*d*sintheta

v = sqrt((4/3)*g*d*sintheta)

v = sqrt((4/3)*9.8*3*sin22)

v = 3.83 m/s

angular speed w(omega) = v/R = 3.83/0.34 = 11.3 rad/s

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from roof to ground

along vertical

initial velocity voy = 0

accelration ay = -g = -9.8 m/s^2

displacement y = -h

from equation of motion

y = voy*t+ (1/2)*ay*t^2

-h = 0 - (1/2)*g*t^2

t = sqrt(2h/g)

along horizopntal

initial velocity vox = v= 3.83 m/s

accelerationa x = 0

displacement = x

x = vox*t + (1/2)*ax*t^2

x = 3.83*sqrt(2*12/9.8) = 6m <<<<----ANSWER

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