A homework assignment including rotation of a rigid body

AI Thread Summary
The discussion revolves around a physics homework problem involving a car decelerating from 30 km/h to a stop over 2 seconds, with a wheel radius of 30 cm. Key calculations include determining the angular acceleration, total angle described by the wheel, number of turns made, and the stopping distance. Participants suggest converting linear speed to rotational velocity and using the given equations to find angular acceleration. The initial poster expresses difficulty in applying the equations but is encouraged to simplify their approach. The conversation emphasizes understanding the relationship between linear and rotational motion for solving the problem effectively.
stipan_relix
Messages
16
Reaction score
1

Homework Statement


The problem is this: A car is evenly slowing down from 30 km/h to 0 in the time of 2 seconds. Radius of its wheels is 30 cm. What is its angular acceleration and what total angle will the wheel describe until it stops? How many turns does the wheel make and what is the length of the path the car travels while stopping?

Homework Equations


\alpha = \frac{\Delta \omega}{\Delta t}\omega = \frac{\phi}{t}\phi= \frac{\alpha}{2}t^2s = vt

The Attempt at a Solution


I honestly tried to use moment of inertia and tangential force equations and I just can't figure it out, every time I find a matching equation, something is missing and I can't solve it. Please help me, at least by some tips if you don't want to do the whole problem. Thanks.
 
Physics news on Phys.org
stipan_relix said:

Homework Statement


The problem is this: A car is evenly slowing down from 30 km/h to 0 in the time of 2 seconds. Radius of its wheels is 30 cm. What is its angular acceleration and what total angle will the wheel describe until it stops? How many turns does the wheel make and what is the length of the path the car travels while stopping?

Homework Equations


\alpha = \frac{\Delta \omega}{\Delta t}\omega = \frac{\phi}{t}\phi= \frac{\alpha}{2}t^2s = vt

The Attempt at a Solution


I honestly tried to use moment of inertia and tangential force equations and I just can't figure it out, every time I find a matching equation, something is missing and I can't solve it. Please help me, at least by some tips if you don't want to do the whole problem. Thanks.
You are over-thinking this. Think about how you can change the linear speed into a rotational velocity of the wheel. Hint: Today is appropriately 3/14/15, using the us convention for the date! From the initial angular velocity and the time to stop, you can calculate alpha, and on you go.
 
Quantum Defect said:
You are over-thinking this. Think about how you can change the linear speed into a rotational velocity of the wheel. Hint: Today is appropriately 3/14/15, using the us convention for the date! From the initial angular velocity and the time to stop, you can calculate alpha, and on you go.
Thank you, I'll try it and report back!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top