Question

In: Physics

A wheel with a weight of 392 N comes off a moving truck and rolls without...

A wheel with a weight of 392 N comes off a moving truck and rolls without slipping along a highway. At the bottom of the hill it is rotating at an angular velocity of 27.7 rad/s. the radius of the wheel is .596 m and its moment of inertia about its rotation axis is .800 MR^2. Friction does work on the wheel as it rolls up the hill to a stop at a height of h about the bottom of the hill; this work has a magnitude of 3464J.

Calculate h

Solutions

Expert Solution

By energy conservation,

KEi + PEi + W = KEf + PEf eq(1)

here, W = work done by friction = Wfr

given Wfr = -3464 J

KE = 0.5*I*w^2 + 0.5*M*v^2

given I = 0.8*M*R^2

R = 0.596 m

weight of wheel = M*g = 392 N

M = 392/9.81 = 39.96 Kg

w = angular velocity = 27.7 rad/sec.

v = w*R = 27.7*0.596 = 16.51 m/sec.

So, KEi = 0.5*0.8*39.96*(0.596)^2*(27.7)^2 + 0.5*39.96*(16.51)^2 = 9802.65 J

KEf = 0

PEi = 0

PEf = M*g*h

here h = height = ??

So, by eq(1),

9802.65 + 0 - 3464 = 0 + 39.96*9.81*h

h = (9802.65 - 3464)/(39.96*9.81)

h = 16.17 m

Please upvote.

By energy conservation,

KEi + PEi + W = KEf + PEf eq(1)

here, W = work done by friction = Wfr

given Wfr = -3464 J

KE = 0.5*I*w^2 + 0.5*M*v^2

given I = 0.8*M*R^2

R = 0.596 m

weight of wheel = M*g = 392 N

M = 392/9.81 = 39.96 Kg

w = angular velocity = 27.7 rad/sec.

v = w*R = 27.7*0.596 = 16.51 m/sec.

So, KEi = 0.5*0.8*39.96*(0.596)^2*(27.7)^2 + 0.5*39.96*(16.51)^2 = 9802.65 J

KEf = 0

PEi = 0

PEf = M*g*h

here h = height = ??

So, by eq(1),

9802.65 + 0 - 3464 = 0 + 39.96*9.81*h

h = (9802.65 - 3464)/(39.96*9.81)

h = 16.17 m

Please upvote.

By energy conservation,

KEi + PEi + W = KEf + PEf eq(1)

here, W = work done by friction = Wfr

given Wfr = -3464 J

KE = 0.5*I*w^2 + 0.5*M*v^2

given I = 0.8*M*R^2

R = 0.596 m

weight of wheel = M*g = 392 N

M = 392/9.81 = 39.96 Kg

w = angular velocity = 27.7 rad/sec.

v = w*R = 27.7*0.596 = 16.51 m/sec.

So, KEi = 0.5*0.8*39.96*(0.596)^2*(27.7)^2 + 0.5*39.96*(16.51)^2 = 9802.65 J

KEf = 0

PEi = 0

PEf = M*g*h

here h = height = ??

So, by eq(1),

9802.65 + 0 - 3464 = 0 + 39.96*9.81*h

h = (9802.65 - 3464)/(39.96*9.81)

h = 16.17 m

Please upvote.


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