A 2.0 kg particle moves in a circle of radius 3.5 m. As you look down on the pla

Question

A 2.0 kg particle moves in a circle of radius 3.5 m. As you look down on the plane of its orbit, the particle is initially moving clockwise. If we call the clockwise direction positive, the particle's angular momentum relative to the center of the circle varies with time according to L(t) = 10 N middot m middot s - (4.5 N middot m)t. (a) Find the magnitude and direction of the torque acting on the particle. (b) Find the angular velocity of the particle as a function of time in the form omega (t) = A + Bt. A = rad/s B = rad/s^2

Explanation / Answer

A) Torque = dL/dt (differentiation)
= -4.5

Direction = tnagential (Clockwise)

Iw = 10-4.5t (where I is moment of inertia and w is angular velocity)
mr^2 * w = 10- 4.5t

2*3.5^2*w = 10 - 4.5t

w = 0.408 - 0.184 t

A = 0.408

B = 0.184

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.