The velocity of a 0.5 kg body, for t = 1s to t = 35s, is given by v = 5t^2i + (3
ID: 2218324 • Letter: T
Question
The velocity of a 0.5 kg body, for t = 1s to t = 35s, is given by v = 5t^2i + (3/t)j - tk. For this interval, A) What is the body's angular momentum B.) What is the torque exerted on the body? C.) What is the acceleration of the body? D.) What is the force on the body? E.) Is either angular momentum or linear momentum conserved? F.) At t -> infinity, what happens to the position of the particle? I know that A)angular momentum =? = mvr r = vdt = 5t^3/3 i - 3/t^2 j -t^2/2 k from t = 1 to 35 s put limits in v and r and get the corresponding values put it in angular momentum formula to get it B) Torque = I? I = moment of inertia = mr^2 putting this gives Torque c) acceleration = v^2/r d) force = m*v^2/r My questions is how do I solve these equations? Say for force, I am taking the v vector squared / r vector (the integral of the v vector), how do I do this? You can't divide vectors right?Explanation / Answer
mass of body is m = 0.5 kg time is t = 1s to t = 35s velocity is v = 5t^2i + (3/t)j - tk we substitute t = 1s and t = 35s in velocity expression and find the velocity v = [(vx^2 + vy^2 + vz^2)]^1/2 where vx,vy and vz are velocity components the acceleration is a = dv/dt = 10ti - (3/t^2)j - k we know that a = v^2/r where r is radius or r = v^2/a the angular momentum is L = mvr the force is F = ma the torque is T = ma * r
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