Answer:
The average linear velocity (inches/second) of the golf club is 136.01 inches/second
Explanation:
Given;
length of the club, L = 29 inches
rotation angle, θ = 215⁰
time of motion, t = 0.8 s
The angular speed of the club is calculated as follows;
The average linear velocity (inches/second) of the golf club is calculated as;
v = ωr
v = 4.69 rad/s x 29 inches
v = 136.01 inches/second
Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second
Answer:
The magnitude of the applied torque is
(e) is correct option.
Explanation:
Given that,
Mass of object = 3 kg
Radius of gyration = 0.2 m
Angular acceleration = 0.5 rad/s²
We need to calculate the applied torque
Using formula of torque
Here, I = mk²
Put the value into the formula
Hence, The magnitude of the applied torque is
Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g
+ m₂ g
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
How energy is conserved
Em₀ =
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)