Question

A small object with mass 3.70 kg moves counterclockwise with constant speed 6.10 m/s in a circle of radius 4.80 m centered at the origin. It starts at the point with position vector (4.80i 0 ) m. Then it undergoes an angular displacement of 9.00 rad. (a) What it its position vector? 1 + i) m (b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis? Selectat A° (c) What is its velocity? (d) In what direction is it moving? Select- (e) What is its acceleration? (f) What total force is exerted on the object? Need Help? Talk to a Tutor

Answer 1

As per HOMEWORKLIB RULES only 1Q and 4 sub-parts, please post the other separately,

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Answer 2

Radius = 4.8 and it starts at 0^o

The moment of inertia = mr² = 3.70x4.8² =85.24

(a) x = 4.8cos155 ; y = 4.8sin155

( --4.35 i + 2.02 j) m in quadrant II (3)

(b) 2nd quadrant at 155.6^o

(c) x = rcosωt, y = rsinωt where r = 4.8, ω = V/r = 6.10/4.8 = 1.27rad/s

dx/dt = Vx = -ωrsinωt i = -(1.27x4.8)sin155.6 = -2.518 i

dy/dt = Vy = ωrcosωt j = (1.27x4.8)cos155.6 = -5.551 j

v = (-2.518 i -5.551 j) m/s

(d) counter clockwise

(e)

dVx/dt = ax = -ω² rcosωt i = (-1.27²x4.8)cos155.6 i = 7.05 i

dVy/dt = ax = -ω² rsinωt j = (-1.27²x4.8)sin155.6 j = 3.19 j

a = (7.05 i + 3.19 j) m/s²

(f) mV²/r = (3.7x6.10²)/4.8 = 28.68 N