Help Solving Angular Work Problem: 11676 J?

AI Thread Summary
To stop a 42.0 kg wheel rotating at 250 rpm, the correct moment of inertia is calculated as I = 34.02 kgm² using the formula for a thin hoop. The angular velocity converts to approximately 26.18 rad/s. Using the rotational kinetic energy equation, the work required to stop the wheel is determined to be 23,724.75 J, indicating that a negative result signifies the work done against the wheel's motion. This highlights that energy must be expended to halt the wheel effectively. The initial calculation of -11,676 J was incorrect due to misapplied values.
Punchlinegirl
Messages
221
Reaction score
0
A 42.0 kg wheel, essentially a thin hoop with radius 0.90 m, is rotating at 250 rpm. It must be brought to a stop in 10 s. How much work must be done to stop it.

I used the equation delta K = 1/2 I w^2_f - 1/2 I w^2_i = W and got - 11676 J, but this is not right. Help?
 
Physics news on Phys.org
Punchlinegirl said:
A 42.0 kg wheel, essentially a thin hoop with radius 0.90 m, is rotating at 250 rpm. It must be brought to a stop in 10 s. How much work must be done to stop it.

I used the equation delta K = 1/2 I w^2_f - 1/2 I w^2_i = W and got - 11676 J, but this is not right. Help?
From problem statement:
{Hoop Mass} = (42.0 kg)
{Hoop Radius} = (0.9 m)
{Moment of Inertia for Thin Hoop} = M*R^2
{Frequency of Rotation} = f = (250 RPM) = (4.1667 rev/sec)
{Angular Velocity} = ω = 2*π*f = 2*π*(4.1667) = (26.18 radians/sec)

{Rotational Kinetic Energy} = (1/2)*I*ω^2 =
= (1/2)*(M*R^2)*ω^2 =
= (1/2)*(42.0)*{(0.9)^2}*{(26.18)^2} =
= (11659 J)


~~
 


Firstly, great job on using the correct equation to solve this problem! However, the negative result you got may be due to the fact that you used the wrong values for the moment of inertia (I) and final angular velocity (w_f).

To calculate the moment of inertia for a thin hoop, we use the formula I = mr^2, where m is the mass and r is the radius. So in this case, the moment of inertia should be I = (42.0 kg)(0.9 m)^2 = 34.02 kgm^2.

Next, we need to convert the given angular velocity of 250 rpm to radians per second. Recall that 1 revolution = 2π radians, so 250 rpm = (250/60) rev/s = (25/6) rev/s = (25/6)(2π) rad/s = 26.18 rad/s.

Now, plugging in the correct values into the equation, we get:

delta K = 1/2 (34.02 kgm^2)(0) - 1/2 (34.02 kgm^2)(26.18 rad/s)^2 = - 23,724.75 J

Since the result is still negative, it means that the work done to stop the wheel is actually 23,724.75 J. This makes sense because in order to stop an object's motion, we need to apply a force in the opposite direction, which requires work to be done.

I hope this helps clarify the confusion and provides the correct solution to the problem. Keep up the good work!
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top