Fido 3:08 PM 97% Moodle 6. The graphs show a wave a function of position at two

Question

Fido 3:08 PM 97% Moodle 6. The graphs show a wave a function of position at two different times. The first at t 0, the second at t 5 ms. Assume the wave is propagating in the-x direction at the smallest possible speed consistent with the graphs. a) Write the wavefunction y(x,t) explicitly b) Write a function for velocity as a function of position and time, v(x.y). 7. A 1.50 kg hollow sphere, with radius 10.0 cm, rolls up a ramp without slipping. Its translational speed at the bottom is 2.50 m/s. The incline of the ramp is 12.0° above the horizontal. The rotational inertia of a hollow sphere is a) What vertical height does it reach? b) If a block of the same mass, also starting at 2.50 m/s, could slide up the ramp without friction, would the height reached be greater, less than, or the same as your justify your an n (a)? Briefly 30° 8. A uniformi disk turns 4:40 s about a frictionless spindle. A nonrotating rod of the same mass as the disk and length equal to disk's diameter is dropped onto the freely spinning disk where it sticks to the disk. They then both turn around the spindle, with their centers superimposed. What is the angular speed (in revs/s) of the combination? Rod: I, L-length of rod Disk R radins

Explanation / Answer

6. the wave is propogating in -x direction

hecne the wavefunction is of the format

y(x, t) = Asin(kx + wt + phi)

now, form the given graphs and data

at t = 0, y = 0, and y < 0 for t > 0 ( in the neighbourhood of t = 0)

hence

y(x,0) = 0 = Asin(kx + phi)

for x = 0

phi = pi

hence

y(x,y) = -Asin(kx + wt)

now from the graph A = 8

lambda = 20

hence

k = 2*pi/lambda = 2*pi/20 = pi/10

also, the wave travels 5 units to the left in 5 ms

hence wave speed, v = 5*1000/5 = 1000

hence w/k = 1000

w = 1000k = 100pi

hence

a. y(x,t) = -8sin(pi*x/10 + 100*pi*t)

b. v(x,t) = dy/dt = -800*pi*cos(pi*x/10 + 100*pi*t)

7. m = 1.5 kg

hollow sphere

r = 10 cm

rolls up a ramp without slipping

vi = 2.5 m/s

theta = 12 deg

a. rotational inertia of a hollow sphere = 2mr^2/3 = 0.01 kgm^2

b. let the final height be h

then from conservation of energy

mgh = 0.5mvi^2 + 0.5Iwi^2

wi = vi/r

hence

mgh = 0.5mvi^1 + 0.5*2m*vi^2/3 = 0.5mvi^2(1 + 2/3)

h = 5vi^2/6g = 0.53092082 m

c. for the block which could slide without friciton, it would have reached lesser height, as the ball which is rolling with the same linear speed has more KE on accountof rolling kinetic energy as well, in addition to the linear KE

8. given w = 4.4 rev/s = 8.8*pi rad/s

uniform disc, mass m, diameter = d

non rotating rod of mass m, lenggth = d

let final angular speed be w

then from conservation of angular momentum

0.5md^2*w/4 = (0.5md^2/4 + md^2/12)w'

3*w = (5)w'

w' = 3w/5 = 2.64 rev/s = 5.28*pi rad/s

9. l = 40 cm

m = 0.5 kg

pivot point is at one end of the rod

force F is acting upwards at r = 30 cm

hence from torque balance

F*30 = mg*40/2

a. net torque on this rod = 0

because this rod is in equilibrium

b. rotational moment of inertia about the pivot point = ml^2/3 = 0.02666666666666 kgm^2

c. angular acceleration = alpha = 0

as torque on the rod is 0

10. as the hanging mass is moving upwards, hecne

if the hanging mass is on the right, the tension on the left string is more than the right string, and that is why the hanging mass is moving upwards

for the hanging mass to the left, then the right string has more tension thatn the left strgin

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