Problem 9: A point particle of mass m = 1.1 kg moves according to the position f

Question

Problem 9: A point particle of mass m = 1.1 kg moves according to the position function: r(t-x:4-y -2fk. where t denotes time and x, y, z, a, b, and c are constants such that the exponents are positive integers and the position function has the dimension of length Part (a) We can write the particle's velocity function in the form )-i orj pek. Enter an expression for n in terms of x,y, z,a,b and c Expression Select from the variables below to wre your expression. Note that all variables may not be required , , a, b, d, g, h, j, k, m, n, P, S, t, x Part (b) The particle's velocity function will have the form v)-n+opk Enter an expression for d in terms ofx, y, , a, b, and o Expression d= Select from the variables below to write your expression. Note that all riables may not be required p.. a, b, d, g, h, j, k, m, n, P, S, t, x Part (c) Here is a set of parameter values for the motion of the particle: m-1.1 kg, x-1.5 ms), 2.3 ms·:-0.85 ms, a = 3, b = 4, Calculate the x-component of the particle's angular momentum, in units of kg·m, about the origin at time t = 1 s. Numeric A numeic value is expected and not an expression. Lx- Part (d) Use the same set of parameter values (m-1.1 kg, x-1.5 ms3.y-2.3 m/s. z-0.35 m/s3,a-3, b-4, c-3) to calculate they- component of the particle's angular momentum: in units of kg·m, about the origin at time t = 1 s. Numeric A numeric value is expected and not an expression.

Explanation / Answer

We know that velocity of a particle is

V=dr/dt

where r is the displacment vector of the particle.

So,

differentiating the given position vector with respect to time,

V=(axta-1)i+(bytb-1)j+(cztc-1)k

part (a)

comparing with the given form of velocity

n=ax

part(b)

again comparing with given form,

d=a-1

part c)

now we know that angular momentum of a body of mass m is

m(r*v)..........................(r*v) is the cross product of two vectors.

So here ,

m=1.1,x=1.5,y=2.3,z=0.85,a=3,b=4,c=3

so r=1.5t3i+2.3t4j+0.85t3k

and v=4.5t2i+9.2t3j+2.55t2k

now doing cross product of r and v

13.8t6k-3.825t5j-10.35t6k+5.865t6i+3.825t5j-7.82t6i

simplifying

-1.955t6i+3.45t6k

so angular momentum will be,

m(r*v)

-2.1505t6i+3.795t6k

at t=1 x component will be

-2.1505

part(d)

at t=1

y component will be zero.

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