An 8 m ladder with a mass of 20 kg lies flat on the ground. A cat with a mass of

Question

An 8 m ladder with a mass of 20 kg lies flat on the ground. A cat with a mass of 4 kg sits on the ladder

at a distance of 2 m from the bottom end of the ladder. A painter, not noticing the cat, grabs the top

end of the ladder and pulls straight upward with a force of 250 N. At the instant the top of the ladder

leaves the ground, the ladder experiences an angular acceleration of 1.80 rad/s2 about an axis passing

through the bottom end of the ladder. (The ladder’s center of gravity lies halfway between the top

and bottom ends.)

a. What is the net torque acting on the ladder+cat system?

b. What is the total moment of inertia of the ladder+cat system about the axis of rotation?

c. What is the moment of inertia of the ladder about the axis of rotation?

d. What is the total acceleration the cat experiences immediately after the ladder leaves the ground?

Explanation / Answer

Mass of Ladder, M= 20kg

Length of ladder, L= 8.0m

Mass of cat, m= 4kg

Distance of cat from bottom end of ladder, d= 2m

Force with which top end of ladder is pulled straight upwards, F= 250N

Angular acceleration of ladder, aa= 1.80 rad/s2

a). Then torque on ladder will be due to the applied force, due to the weight of ladder and due to the weight of the cat.

Since ladder is uniform hence its weight will act at exactly halfway between the ladder. Then net torque will be

T= FLsin90+Mg(L/2)sin270+mgdsin270 (calculating angle clockwise b/w force and distance)

using given values in above,

T= (250)(8)+ 20(9.8)(8/2)(-1)+ (4)(9.8)(2)(-1)

T= 2000-784-78.4= 1137.6 Nm (ANS)..........................(1)

b). We also know that torque is given by

T= Itotaa

where Itot= Total moment of inertia of ladder-cat system

then using equation 1 and given value of angular acceleration,

Itot= T/aa= 1137.6/1.80= 632 kgm2 (ANS)............................(2)

c). Axis of rotation of the system is bottom end of the ladder. Then we know that

Total Moment of Inertia= Moment of Inertia of ladder+Moment of Inertia of Cat

using equation 2 and other given values in above,

632= Iladder+ md2

632= Iladder+ (4)(2)2

Iladder= 632-16= 616 kgm2 (ANS)

d). Angular acceleration of ladder-cat system, aa= 1.80 rad/s2

Distance of cat from axis of rotation, d= 2m

Then total linear acceleration experienced by cat will be

al= aad= (1.80)(2)= 3.6 m/s2   (ANS)

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