What is the angular momentum of a 3.7kg uniform cylindrical grinding wheel of radius 25cm when rotating at 1500rpm?
How much torque(in magnitude) is required to stop it in 5.0s?
Could you explain this in detail, I don't really understand torque and momentum well.
You need to separate linear motion from rotational.
Read in two columns
mass kg...........momentofinertia kg*m^2
displacement m...angular displacement rad
velocity m/s ....ang velocity rad/s^2
acceleration m/s^2.ang acc rad/s^2
and finally, memorize this
tangential velocity on a curve= radius*ang velocity
Now your question.
ang momentum=I*w
(that is ang momentum is momentofinertia * angular velocity)
So you need I for a solid disk, make you a convenient card with common shapes..http://en.wikipedia.org/wiki/List_of_moments_of_inertia
Using thin disk model, I= mr^2/2
ang momentum= mr^2/2 * 1500rev/min*1min/60sec*2PIrad/rev
b) Torque= I*angacceleration= I*(wf-wi)/time= mr^2/2(-1500*2PI/60)
To calculate the angular momentum of the grinding wheel, we can use the formula:
Angular momentum (L) = moment of inertia (I) x angular velocity (ω)
The moment of inertia of a uniform cylindrical object is given by:
Moment of inertia (I) = 0.5 x mass (m) x radius squared (r^2)
Given:
- Mass of the grinding wheel (m) = 3.7 kg
- Radius of the grinding wheel (r) = 25 cm = 0.25 m
- Angular velocity (ω) = 1500 rpm = 1500 revolutions per minute
First, let's convert the angular velocity from rpm to rad/s:
1 revolution = 2π radians
1 minute = 60 seconds
Angular velocity (ω) = 1500 rpm x (2π/60) rad/s
= 1500 x (2π/60) rad/s
= 1500π/30 rad/s
= 50π rad/s
Now, let's calculate the moment of inertia (I):
Moment of inertia (I) = 0.5 x m x r^2
= 0.5 x 3.7 kg x (0.25 m)^2
= 0.5 x 3.7 kg x 0.0625 m^2
= 0.115625 kg.m^2
Finally, let's calculate the angular momentum (L):
Angular momentum (L) = I x ω
= 0.115625 kg.m^2 x 50π rad/s
= 5.78125π kg.m^2/s
Therefore, the angular momentum of the grinding wheel is 5.78125π kg.m^2/s.
Now let's move on to calculating the torque required to stop the wheel in 5.0 seconds.
Torque (τ) = change in angular momentum (ΔL) / time (t)
Given:
- Time (t) = 5.0 s
To calculate the change in angular momentum (ΔL), we need to subtract the final angular momentum (Lf) from the initial angular momentum (Li). Since the wheel is being stopped, the final angular momentum will be zero, as there is no more rotation.
ΔL = Lf - Li
= 0 - 5.78125π kg.m^2/s
= -5.78125π kg.m^2/s
Now let's calculate the torque (τ):
Torque (τ) = ΔL / t
= (-5.78125π kg.m^2/s) / 5.0 s
= -1.15625π kg.m^2/s^2
Since torque is a vector quantity, the magnitude of the torque required to stop the grinding wheel in 5.0 seconds is approximately 1.15625π kg.m^2/s^2.
I hope this explanation helps! Let me know if you have any further questions.
To find the angular momentum of the grinding wheel, we need to use the formula:
Angular momentum (L) = moment of inertia (I) × angular velocity (ω)
The moment of inertia of a uniform cylindrical object can be calculated using the formula:
Moment of inertia (I) = (1/2) × mass (m) × radius^2
Given:
Mass of the grinding wheel (m) = 3.7 kg
Radius of the grinding wheel (r) = 25 cm = 0.25 m
Angular velocity (ω) = 1500 rpm
First, we need to convert the angular velocity from rpm to radians per second (rad/s). Since 1 rpm = 2π radians per minute, we have:
Angular velocity (ω) = 1500 rpm × (2π rad/1 min) × (1 min/60 s) = 1500 × 2π / 60 rad/s ≈ 157.08 rad/s (rounded to 2 decimal places)
Now we can calculate the moment of inertia:
Moment of inertia (I) = (1/2) × mass (m) × radius^2
= (1/2) × 3.7 kg × (0.25 m)^2
= 0.46 kg⋅m^2 (rounded to 2 decimal places)
Finally, we can find the angular momentum:
Angular momentum (L) = moment of inertia (I) × angular velocity (ω)
= 0.46 kg⋅m^2 × 157.08 rad/s
= 72.30 kg⋅m^2/s (rounded to 2 decimal places)
So, the angular momentum of the grinding wheel is approximately 72.30 kg⋅m^2/s.
Moving on to the second part of the question, we need to find the torque required to stop the grinding wheel in 5.0 seconds.
Torque (τ) = change in angular momentum (ΔL) / time interval (Δt)
Since we want to stop the wheel, the change in angular momentum (ΔL) is equal to the negative of the initial angular momentum (L). Therefore:
ΔL = -L = -72.30 kg⋅m^2/s
The time interval Δt is given as 5.0 seconds.
Plugging these values into the equation for torque, we have:
Torque (τ) = ΔL / Δt = (-72.30 kg⋅m^2/s) / (5.0 s) = -14.46 N⋅m (rounded to 2 decimal places)
The magnitude of torque is equal to 14.46 N⋅m.
So, approximately 14.46 Newton-meters of torque are required to stop the grinding wheel in 5.0 seconds.