A 1.5 kg , 20-cm-diameter turntable rotates at 110 rpm on frictionless bearings. Two 550 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick.
What is the turntable's angular velocity, in rpm, just after this event?
To find the turntable's angular velocity just after the event, we can first calculate the initial angular momentum of the system before the blocks hit the turntable. The angular momentum is given by the equation:
L = Iω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia of the turntable can be calculated using the equation:
I = (1/2)MR²
Where M is the mass of the turntable and R is the radius.
Given:
- Mass of the turntable (M) = 1.5 kg
- Diameter of the turntable = 20 cm
- Radius of the turntable (R) = 10 cm = 0.1 m
Substituting the values into the equation, we can calculate the moment of inertia:
I = (1/2)(1.5 kg)(0.1 m)²
Simplifying the equation gives:
I = 0.075 kg·m²
Before the blocks hit the turntable, its angular velocity is given as 110 rpm (revolutions per minute). To convert this to radians per second (rad/s), we can use the conversion factor:
1 revolution = 2π radians
So, 110 rpm can be converted to radians per second as:
ω_initial = (110 rpm) * (2π rad/1 min) * (1 min/60 s)
Simplifying the equation gives:
ω_initial = (11π/3) rad/s
Since the blocks stick to the turntable and the moment of inertia does not change after the event, the final angular momentum can be obtained by summing the initial angular momentum of the turntable and the blocks.
The masses of the blocks are given as 550 g each, so the total mass of the blocks is:
m_blocks = 550 g + 550 g = 1100 g = 1.1 kg
The distance between the center of the turntable and the blocks is equal to the radius of the turntable, which is 0.1 m.
The angular momentum of the blocks can be calculated using the equation:
L_blocks = m_blocks * v_blocks * R
Where v_blocks is the velocity of the blocks.
The velocity of the blocks can be calculated using the equation:
v_blocks = ω_initial * R
Substituting the values, we can calculate the angular momentum of the blocks:
L_blocks = (1.1 kg) * (11π/3 rad/s) * (0.1 m)
Simplifying the equation gives:
L_blocks ≈ 1.138 kg·m²/s
Adding the angular momentum of the blocks and the turntable gives the final angular momentum:
L_final = L_blocks + L_turntable
L_final = 1.138 kg·m²/s + 0.075 kg·m²/s
L_final ≈ 1.213 kg·m²/s
Finally, we can calculate the final angular velocity (ω_final) by rearranging the equation for angular momentum:
L_final = I * ω_final
ω_final = L_final / I
Substituting the values, we can calculate the final angular velocity:
ω_final ≈ (1.213 kg·m²/s) / (0.075 kg·m²)
Simplifying the equation gives:
ω_final ≈ 16.173 rad/s
To convert this to rpm, the final angular velocity can be multiplied by the conversion factor:
1 rev = 2π rad
ω_final_rpm = (16.173 rad/s) * (1 rev/2π rad) * (60 s/1 min)
Simplifying the equation gives:
ω_final_rpm ≈ 154.5 rpm
Therefore, the turntable's angular velocity, just after the event, is approximately 154.5 rpm.
To find the turntable's angular velocity just after the event, we can use the principle of conservation of angular momentum. The angular momentum before the blocks hit the turntable is equal to the angular momentum after the blocks stick to the turntable.
The angular momentum, L, of a rotating object is given by the equation:
L = I ω
where I is the moment of inertia and ω is the angular velocity.
The moment of inertia of a solid disc rotating about its axis is given by:
I = (1/2) M R^2
where M is the mass of the disc and R is its radius.
Given that the turntable has a mass of 1.5 kg and a diameter of 20 cm (which means a radius of 10 cm or 0.1 m), we can calculate its moment of inertia:
I = (1/2) * 1.5 kg * (0.1 m)^2 = 0.075 kg·m^2
The total angular momentum before the blocks hit the turntable is zero, as there are no external torques acting on the system. After the blocks stick to the turntable, the moment of inertia of the system increases to include the masses of the blocks. The combined mass of the turntable and the blocks is:
M_total = M_turntable + M_block1 + M_block2
= 1.5 kg + 0.55 kg + 0.55 kg
= 2.6 kg
The moment of inertia of the system can be calculated using the formula mentioned earlier:
I_total = (1/2) * M_total * R^2
= (1/2) * 2.6 kg * (0.1 m)^2
= 0.013 kg·m^2
The total angular momentum after the blocks hit the turntable is the product of the combined moment of inertia and the final angular velocity, ω_final.
0 = I_total * ω_final
Solving for ω_final, we get:
ω_final = 0 / I_total
= 0 rad/s
To convert from radians per second to revolutions per minute (rpm), we need to multiply by a conversion factor. Since 1 revolution is equal to 2π radians, and there are 60 seconds in a minute, the conversion factor is:
1 rad/s = (60 s / 2π radians) * (1 revolution / 1 radian) * (1 minute / 60 s)
= 30 / π revolutions/minute
≈ 9.55 revolutions/minute
Therefore, the turntable's angular velocity just after the event is approximately 0 rpm.
www.slader.com/discussion/question/a-20-kg-20-cm-diameter-turntable-rotates-at-100-rpm-on-frictionless-bearings-two-500-g-blocks-fall-from-above-hit-the-turntable-simultaneous-7525d7c6/
^Copy-paste that link into your search bar. It's a similar question to yours but instead of a 1.5 kg block its a 2 kg block. Instead of 110 rpm, its 100 rpm and instead of two 550 g blocks its two 500 g.
Switch the numbers and solve.
Good luck!