Fill in the blanks A 1.31 kg toy car moves along an x axis with a velocity given

Question

 Fill in the blanks

A 1.31 kg toy car moves along an x axis with a velocity given by v = -8.00t3i m/s, with t in seconds. For t > 0, what are (a) the angular momentum L of the car and (b) the torque on the car, both calculated about the origin? What are (c)L and (d) about the point (8.43 m, 5.55 m, 0)? What are (e)L and (f) about the point (8.43 m, -5.55 m, 0)?

(a) (              kg·m2/s)i + (                  kg·m2/s)j + (                       kg·m2/s)k
(b) (               N·m)i + (                N·m)j + (                N·m)k
(c) (                 kg·m2/s)i + (                   kg·m2/s)j + (                 *t3 kg·m2/s)k
(d) (                    N·m)i + (                       N·m)j + (               *t2 N·m)k
(e) (                 kg·m2/s)i + (                 kg·m2/s)j + (                      *t3 kg·m2/s)k
(f) (                     N·m)i + (                  N·m)j + (                *t2 N·m)k

Explanation / Answer

a) (0kg·m2/s)i + (0kg·m2/s)j + ( 0kg·m2/s)k because cross product of (rxp)
both are in i direction so it is zero
b)similarly
(b) (0 N·m)i + (0N·m)j + (0N·m)k
(c) r=(2t4-8.43)i -5.55j

so L=rxp p=(mV)

( 0 kg·m2/s)i + ( 0kg·m2/s)j - (58.164*t3kg·m2/s)k

(d)T=rxf f=ma a=24t^2

( 0 N·m)i + ( 0 N·m)j - (174.492*t2N·m)k

(e) r=(2t4-8.43)i +5.55j

so L=rxp p=(mV)

( 0 kg·m2/s)i + ( 0kg·m2/s)j + (58.164*t3kg·m2/s)k

(f)T=rxf f=ma a=24t^2

( 0 N·m)i + ( 0 N·m)j + (174.492*t2N·m)k

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