How fast, in rpm, would a 5.6 kg, 25-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s
1 answer:
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Answer:
τ = 132.773 lb/in² = 132.773 psi
Explanation:
b = 12 in
F = 60 lb
D = 3.90 in (outer diameter) ⇒ R = D/2 = 3.90 in/2 = 1.95 in
d = 3.65 in (inner diameter) ⇒ r = d/2 = 3.65 in/2 = 1.825 in
We can see the pic shown in order to understand the question.
Then we get
Mt = b*F*Sin 30°
⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in
Now we find ωt as follows
ωt = π*(R⁴ - r⁴)/(2R)
⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)
⇒ ωt = 2.7114 in³
then the principal stresses in the pipe at point A is
τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)
⇒ τ = 132.773 lb/in² = 132.773 psi
Answer:
Explanation:
<u>Motion With Constant Acceleration </u>
It's a type of motion in which the velocity of an object changes uniformly over time.
The equation that describes the change of velocities is:
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
Solving the equation for a:
The ball starts at rest (vo=0) and rolls down an inclined plane that makes it reach a speed of vf=7.5 m/s in t=3 seconds.
The acceleration is: