Answer:
0.20kg-m^2
Explanation:
Let the linear velocity of the rope(=of pulley) is v m/s
Using kinematic equation
=> v = u + at
=>v = 0 + 4.9a
=>v = 4.9a ------------ eq1
By v^2 = u^2 + 2as
=>v^2 = 0 + 2 x v/4.9 x 1.2
=>4.9v^2 - 2.4v = 0
=>v(4.9v - 2.4) = 0
=>v = 2.4/4.9 = 0.49 m/s
Thus by v = r x omega
=>omega = v/r = 0.49/0.02 = 24.49 rad/sec
BY W = F x s = 50 x 1.2 = 60 J
=>KE(rotational) = W = 1/2 x I x omega^2
=>60 = 1/2 x I x (24.49)^2
=>I = 0.20 kg-m^2
Answer:
a) 4.49Hz
b) 0.536kg
c) 2.57s
Explanation:
This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

for some time t you have:
x=0.134m
v=-12.1m/s
a=-107m/s^2
If you divide the first equation and the third equation, you can calculate w:

with this value you can compute the frequency:
a)

b)
the mass of the block is given by the formula:

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

Finally, the amplitude is:

Answer:
calm down please its not that serious maybe no one saw it yet
Explanation:
we know that center of mass is given as
r = (m₁
+ m₂
)/(m₁ + m₂)
taking derivative both side relative to "t"
dr/dt = (m₁ d
/dt + m₂ d
/dt)/(m₁ + m₂)
v = (m₁
+ m₂
)/(m₁ + m₂)
taking derivative again relative to "t" both side
dv/dt = (m₁ d
/dt + m₂ d
/dt)/(m₁ + m₂)
a= (m₁
+ m₂
)/(m₁ + m₂)
They communicate their result to the scientific community- so to speak