Question

IA weight of mass 6.89 kg is suspended by a string of length 0.849 m, and set into motion along circular horizontal path (see figure). The angle \theta? of the string with respect to vertical is 5.61^o?o??. What is the period of the circular motion? I can not uplaod the image!

Also I got 11 for this problem but apparently that is wrong: A mass, m?1??=24.9 kg mass is placed on a frictionless ramp which is inclined 43.5^\circ???? above horizontal. It is connected to a second mass, m?2??, by a strong rope which runs over a pulley at the apex of the ramp, so that the second mass is suspended in the air next to the ramp, as shown in the figure. Calculate the value of m?2?? necessary so that the first mass accelerates up the incline at rate of 1.66 m/s^2???.

I did 24.9*sin(43.5)-1.66/[1.66+9.8]

Answer 1

in vertical,

T cos5.61 = m g = 6.89 x 9.8

T = (6.89 x 9.8)/cos5.61

in horizontal,

T sin5.61 = m v^2/ r

(6.89 x 9.8 / cos5.61) = 6.89 v^2 / (0.849 x sin5.61)

v = 0.904 m/s

anr r = 0.849 sin5.61 = 0.083 m

T = 2 pi r / v = 0.577 sec ......Ans

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on m1:

T - m1 g sin43.5 = m1 a
on m2:

m2 g - T = m2 a

9.8 m2 - (24.9 x 9.8 x sin43.5) = (24.9 x 1.66) + 1.66 m2

m2 = ((24.9 x 9.8 x sin43.5) + (24.9 x 1.66))/(1.66 + 9.8)

m2 = 18.3 kg