Problem 1 (30 Points) A wheel W has mass mw 2 kg and rolls on the truck T of mas

Question

Problem 1 (30 Points) A wheel W has mass mw 2 kg and rolls on the truck T of mass m 3 kg shown below. Ihe mass of the truck wheels may be neglected in this problem. The surface on which the truck rolls is inclined 30° from the horizontal. (a) (10 Points) Derive an expression for the angular acceleration of the wheel W, @wjF, in terms of the accelerations of the mass centers of W and T, äc and ig respectively. (b) (20 Points) Find the force P such that the mass center of W (point C) does not move in the xy plane. Moment of Inertia of a wheel: Imr2 mw 2 kg R 2m k9 1T 30°

Explanation / Answer

rw=2m

mw=2kg

mT=3kg

moment inertia of wheel is given as=1/2mr2

the givem moment of inertia is about center of wheel then to find the moment of inertia about point o we have to apply parallel axis theorem

so moment of inertia about point o=1/2mr2+mr2=3/2mr2=12kg.m2

(1) torque about point o=mwg×sin30o×rw=19.62N.m

we know that torque=moment of inertie×angular accilaretion

than angular accileration =19.62/12=1.635rad/s2

(2) if we consider the wheel sepreatly then

mwg sin30o=mw×aw

aw=g sin30o=4.905 m/s2

then for belencing aT should be equelel to aw

aT=aw=4.905m/s2

then

(mT+mw)×4.905= p - (mwg ssin30o + mTg sinin30o)

5×4.905=p - ( 2×9.81sin30o+ 3×9.81 sin30o)

on solving we get

p=49.05N

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