Respuesta :
Answer:
2.38 m/s, 4.31 m/s, lower
Explanation:
a)
Initial energy = final energy
½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²
Since the ball is rolling without slipping, ω = v / r.
For a hollow sphere, I = ⅔ m r².
½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²
½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²
⅚ m v₀² = mgh + ⅚ m v₁²
⅚ v₀² = gh + ⅚ v₁²
v₀² = 1.2gh + v₁²
v₁ = √(v₀² − 1.2gh)
Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:
v₁ = √((4.03)² − 1.2 (9.80) (0.900))
v₁ ≈ 2.38 m/s
At the top of the loop, the sum of the forces in the radial direction is:
∑F = ma
W + N = m v² / R
N = m v² / R - mg
N = m (v² / R - g)
Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:
N = m ((2.38)² / 0.450 - 9.80)
N = 2.77m
N ≥ 0, so the ball stays on the track.
b)
Initial energy = final energy
Borrowing from part a):
v₂ = √(v₀² − 1.2gh)
This time, h = -0.200 m:
v₂ = √((4.03)² − 1.2 (9.80) (-0.200))
v₂ ≈ 4.31 m/s
c)
Without the rotational energy:
½ m v₀² = mgh + ½ m v₁²
½ v₀² = gh + ½ v₁²
v₀² = 2gh + v₁²
v₁ = √(v₀² - 2gh)
This is less than v₁ we calculated earlier.
Refer the below Solution for better understanding.
Given :
Speed = 4.03 m/sec
Vertical circular loop of 90 cm diameter.
Solution :
a)
Initial energy = Final Energy
[tex]\rm \dfrac{1}{2}mv_0^2 + \dfrac{1}{2}I\omega_0^2 = mgh + \dfrac{1}{2}mv_1^2 + \dfrac{1}{2}I\omega_1^2[/tex]
here,
[tex]\rm \omega = \dfrac{v}{r}[/tex]
for a hollow sphere,
[tex]\rm I = \dfrac{2}{3}mr^2[/tex]
[tex]\rm \dfrac{1}{2}mv_0^2 + \dfrac{1}{2}\times\dfrac{2}{3}mr^2(\dfrac{v_0}{r})^2 = mgh + \dfrac{1}{2}mv_1^2 + \dfrac{1}{2}\times\dfrac{2}{3}mr^2(\dfrac{v_1}{r})^2[/tex]
by further solving above equation,
[tex]\rm v_1=\sqrt{v_0^2-1.2gh}[/tex] --- (1)
Now put the values of [tex]\rm v_0 , \;g\;and\;h[/tex] in equation (1),
[tex]\rm v_1 = \sqrt{4.03^2-1.2(9.8)(0.9)}[/tex]
[tex]\rm v_1 = 2.38 \; m/sec[/tex]
Now,
F = ma
[tex]\rm mg + N = \dfrac{mv_1^2}{R}[/tex]
[tex]\rm N = m(\dfrac{v_1^2}{R}-g)[/tex] --- (2)
Now put the values of R, g, m and [tex]\rm v_1[/tex] in equation (2) we get,
N = 2.77m
[tex]\rm N\geq 0[/tex]
ball stays on the track.
b) To find the speed of the ball as it leaves the track,
[tex]\rm v_2=\sqrt{v_0^2-1.2gh}[/tex] ---- (3)
put h = -0.2m in equation (3)
[tex]\rm v_2=\sqrt{4.03^2-1.2(9.8)(-0.2)}[/tex]
[tex]\rm v_2=4.31\;m/sec[/tex]
c) Again, but without rotational energy
Initial energy = Final energy
[tex]\rm \dfrac{1}{2}mv_0^2 = mgh + \dfrac{1}{2}mv_1^2[/tex]
by further solving the above equation we get,
[tex]\rm v_1 = \sqrt{v_0^2-2gh}[/tex] and this is less than [tex]\rm v_1[/tex] we calculated earlier.
For more information, refer the link given below
https://brainly.com/question/7590442?referrer=searchResults
