Question

3. 18 points] Sand moves without slipping at speed vo down a conveyer belt tilted at 15°, and then enters a barrel at h 3.0 m below the end of the conveyer belt, as shown below. If the barrel's diameter is D 90 cm and the barrel is located at d 2.0 m from the conveyer belt, what is the range of speeds vo for which the sand will enter the barrel? 15" d D

Answer 1

3)Given,

theta = 15 deg ; h = 3 m ; D = 90 cm ; d = 2 m

the x and y component of intial velocity will be:

vx = v0 cos15 ; vy = v0 sin15

We know that range D is given by:

d = vx t => t = d/vx

t = 0.9/(v0 cos15)

from eqn of motion, in y direction

S = ut + 1/2 at^2

-3 = v0 sin15 t - 1/2 x 9.8 x t^2

putting the vaue of t in above eqn

-3 = v0 x sin15 x 0.9/v0 cos15 - 4.9 x 0.9^2/v0^2 cos^15

-3 = 0.24 - 4.25/v0^2

4.25/v0^2 = 0.24 + 3 = 3.24

v0 = sqrt (4.25/3.24) = 1.15 m/s

Hence, range is v0 = +-1.15 m/s