Answer: 3.31s
Explanation:
Given that,
Radius = diameter/2 = 1/2 =0.5m
U = 1.9m/s
V=0m/s
The elevator stopped as the cylinder made one complete oscillation.
Vertical distance travelled by the elevator is the same as the circumference of the cylinder.
Hence, s = 2πr
S = 2×3.142×0.5
S= 3.142m
V2=u2+2as
0 = 1.9square + 2×ax3.142
0 = 3.61 + 6.284a
-3.61 = 6.284a
a = 3.61/6.284
a = -0.574m/s2
The time taken for the elevator to stop
V= u+at
0 = 1.9+ ( -0.574 )t
0 = 1.9-0.574t
-1.9= -0.574t
t = 1.9/0.574
t = 3.31 s
In the formulary used above,
V is the final velocity,
U is the initial velocity
A is the acceleration
T is the time
Answer:
Explanation:
Given:
theta = 1 rev
= 2pi rads
S = r × theta
= 0.5 × 2pi
= 3.142 rad
Vo = 1.9 m/s
D = 1 m
R = 0.5 m
w × r = V
wo = 1.9/0.5
= 3.8 rad/s
wf = 0 rad/s
Using angular equations of motion,
wf^2 = wo^2 + 2aoS
ao = 3.8^2/2 × 6.284
= 1.15 rad/s^3
wf = wo + (ao × t)
0 = 3.8 + 1.15t
t = 3.3 s.
