A 7.0 kg bowling ball has a moment of inertia of 2.8x10-2 kg m2, and a radius of 0.10 m. If it rolls down the lane at an angular
1 answer:
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Answer:
The initial velocity is 0.5114 m/s or 511.4 mm/s
Explanation:
Let the initial velocity be 'v'.
Given:
Mass of the ball (m) = 130 g = 0.130 kg [ 1 g = 0.001 kg]
Initial height of the ball (h₁) = 1.4 mm = 0.0014 m [ 1 mm = 0.001 m]
Final height of the ball (h₂) = 15 mm = 0.015 m
Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.
Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.
Change in kinetic energy is given as:
As it just touches the 15 mm high roof, the final velocity will be zero. So,
.
Now, the change in kinetic energy is equal to:
Change in gravitational potential energy = Final PE - Initial PE
So,
[ g = 9.8 m/s²]
Now, Change in KE = Change in PE
Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s
Answer:
The tension in each half of the rope, is approximately 4,908.8 N
Explanation:
The mass of the acrobat, m = 60 kg
The length of the rope, l = 10 m
The extent by which the center dropped = 30 cm = 0.3 m
Let, 'T' represent the tension in each half of the rope
Weight, W = Mass, m × The acceleration due to gravity, g
∴ W = m × g
The acceleration due to gravity, g ≈ 9.8 m/s²
∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N
The angle the dropped rope makes with the horizontal, θ is given as follows;
θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°
At equilibrium, the sum of vertical forces, = 0
The vertical component of the tension, , in each half of the rope is given as follows;
= T × sin(θ)
∴ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)
Plugging in the values, with θ = arctan(0.06) for accuracy, we get;
588 N + 2 × T × sin(arctan(0.06) = 0
∴ 2 × -T × sin(arctan(0.06) = 588 N
-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N
The tension in each half of the rope, T ≈ 4,908.8 N.
Answer:
(a). The value of d is 0.056 cm and 1.496 cm.
(b). The time period is 1.35 sec.
Explanation:
Given that,
Length = 50.00 cm
Time period = 2.50 s
Time period of pendulum is defined as the time for one complete cycle.
The period depends on the length of the pendulum.
Using formula of time period
Where, I = moment of inertia
We need to calculate the value of d
Using parallel theorem of moment of inertia
For a meter stick mass m , the rotational inertia about it's center of mass
Where, L = 1 m
Put the value into the formula of time period
Multiplying both sides by d
tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]
Put the value of T, L and g into the formula
The value of d is 0.056 cm and 1.496 cm.
(b). Given that,
L = 50-5 = 45 cm
We need to calculate the time period
Using formula of period
Put the value into the formula
Hence, (a). The value of d is 0.056 cm and 1.496 cm.
(b). The time period is 1.35 sec.