You are Here: Home >< Physics

# Torque, g/cm? watch

1. I have a project to design an RC car. It weighs 1.45KG and needs to travel at 2m/s .

I've worked out force needed which is 2.9kilo Newtons or 2900N .

But unfortunately my lecturer isn't very good at explaining ( or i cant take it in (I'm Dyslexic)) so i dont really know what to do with the force worked out.

I've managed to find online that Torque = Force (lb) x Distance (ft)

So i converted my numbers to into the equation using grams and cm. The reason i used grams and cm is because i need to find a motor with the right out put of torque ( they are measured in g/cm)

my equation now looks like this
Torque = 1450x200
= 290000 g/cm

I think that is wrong because a motor i am looking at is around 10 g/cm

SOMEONE HELP ME PLEASE

If you don't know, any advice would be helpful,
Thanks,
Matt
2. (Original post by Shanly)
I have a project to design an RC car. It weighs 1.45KG and needs to travel at 2m/s .

I've worked out force needed which is 2.9kilo Newtons or 2900N .

But unfortunately my lecturer isn't very good at explaining ( or i cant take it in (I'm Dyslexic)) so i dont really know what to do with the force worked out.

I've managed to find online that Torque = Force (lb) x Distance (ft)

So i converted my numbers to into the equation using grams and cm. The reason i used grams and cm is because i need to find a motor with the right out put of torque ( they are measured in g/cm)

my equation now looks like this
Torque = 1450x200
= 290000 g/cm

I think that is wrong because a motor i am looking at is around 10 g/cm

SOMEONE HELP ME PLEASE

If you don't know, any advice would be helpful,
Thanks,
Matt
Let's start at the beginning because this looks a bit confused.

Firstly, can you tell us how you came to the conclusion that your car needs a force of 2900N.
How did you arrive at this from its mass and its need to travel at 2m/s?
If you have just multiplied 1.45 x 2 to get 2.9kg, why is that?
Secondly, 2.9kg gives a force (weight) of 29N not 2900N (mg = mass x 10m/s/s)
3. (Original post by Stonebridge)
Let's start at the beginning because this looks a bit confused.

Firstly, can you tell us how you came to the conclusion that your car needs a force of 2900N.
How did you arrive at this from its mass and its need to travel at 2m/s?
If you have just multiplied 1.45 x 2 to get 2.9kg, why is that?
Secondly, 2.9kg gives a force (weight) of 29N not 2900N (mg = mass x 10m/s/s)

Well from what i've learnt, to find out how much force is used to move a box along the ground is f=ma ( although reading this i have realised that gravity is the acceleration. and my lecturer has informed me their is no friction coefficient either)
So originally i did this F=ma so 1.45kg x 2m/s/s =2900N
which should be 1.45 x9.8 = 14.21N
I need the car to move at 2m per second,

Am i right in thinking that this equation gives me the right force needed to move at a constant speed? ---> Fa=(1.45 x 2) + (0) (1.45 X 9.8)
=17.11N

Im getting really confused and doubting what i know
4. (Original post by Shanly)
Well from what i've learnt, to find out how much force is used to move a box along the ground is f=ma ( although reading this i have realised that gravity is the acceleration. and my lecturer has informed me their is no friction coefficient either)
So originally i did this F=ma so 1.45kg x 2m/s/s =2900N
which should be 1.45 x9.8 = 14.21N
I need the car to move at 2m per second,

Am i right in thinking that this equation gives me the right force needed to move at a constant speed? ---> Fa=(1.45 x 2) + (0) (1.45 X 9.8)
=17.11N

Im getting really confused and doubting what i know
You need to distinguish between speed (or velocity) and acceleration.
You stated the car needed to "travel at 2m/s"
2m/s is a uniform speed. I now assume you meant that the car needs to accelerate at 2m/s/s.
In this case you can use F=ma to find the force.
The mass is 1.45kg so the force required is 1.45 x 2 = 2.9N

There is no need to multiply this by g
That formula gives you the force required in newton.

Next we come to torque.
Torque is force times the perpendicular distance the force acts from the pivot or point of rotation. Its unit is newton metre (or Nm)
In the case of a car, the engine produces torque which is transmitted to the wheels and translates to force through the wheel and axle where
the force applied at the ground x the radius of the wheel = torque.
5. Right, so i need a constant force of 2.9N to accelerate the car at 2/s/s , but then i need a constant speed of 2m/s. I thought we used f=ma to find out the force needed for a constant speed
f= (1.45x9.81) but my lecturer has told me that their is no fiction coefficient (because of wheels) which makes the calculation have a ridiculous outcome.

Right so if my force applied at the wheels was 2.9N and my wheels are 10cm in diameter the formula is Torque = 2.9 x 0.05 = 0.145nm

I think my problem i had with finding the torque was that it wasn't explained to me what was the Distance ( radius ).
6. (Original post by Shanly)
Right so if my force applied at the wheels was 2.9N and my wheels are 10cm in diameter the formula is Torque = 2.9 x 0.05 = 0.145nm
got the torque side of things covered then.

2.9/9.8 = 0.2959 kg x1000 = 295.9184g

0.05 m x100 = 5cm
Torque = 295.9184 x 5 = 1479.592gcm
7. (Original post by Shanly)
Right, so i need a constant force of 2.9N to accelerate the car at 2/s/s , but then i need a constant speed of 2m/s. I thought we used f=ma to find out the force needed for a constant speed
f= (1.45x9.81) but my lecturer has told me that their is no fiction coefficient (because of wheels) which makes the calculation have a ridiculous outcome.

Right so if my force applied at the wheels was 2.9N and my wheels are 10cm in diameter the formula is Torque = 2.9 x 0.05 = 0.145nm

I think my problem i had with finding the torque was that it wasn't explained to me what was the Distance ( radius ).
If there is no friction you don't need a force to keep an object moving at constant speed. (Newton's 1st Law).
In reality there is friction (ground and air) so a car needs to develop power to keep it moving at constant speed against air and road resistance.
F=ma is used for acceleration.

Yes
If your wheels are 10cm in diameter then

0.05 x 2.9 = torque required
8. (Original post by Shanly)
got the torque side of things covered then.

2.9/9.8 = 0.2959 kg x1000 = 295.9184g

0.05 m x100 = 5cm
Torque = 295.9184 x 5 = 1479.592gcm
Seems ok if this is converting force (weight) to mass in gram via F=mg
9. (Original post by Stonebridge)
If there is no friction you don't need a force to keep an object moving at constant speed. (Newton's 1st Law).
In reality there is friction (ground and air) so a car needs to develop power to keep it moving at constant speed against air and road resistance.
F=ma is used for acceleration.

Yes but if their is no friction, and taking newtons law into consideration, surely that means my car will be accelerating at 2m/s/s. Does that not mean that after 2 seconds it will be moving at a speed of 4m/s/s, then 3 seconds at 8m/s/s and so forth ?

Sorry for all the questions

And thanks again for everything
10. (Original post by Shanly)
Yes but if their is no friction, and taking newtons law into consideration, surely that means my car will be accelerating at 2m/s/s. Does that not mean that after 2 seconds it will be moving at a speed of 4m/s/s, then 3 seconds at 8m/s/s and so forth ?

Sorry for all the questions

And thanks again for everything
After 1s it's 2m/s
After 2s it's 4m/s
after 3s it's 6m/s
after 4s it's 8m/s

The formula is v = u + at
where u is the initial velocity (here zero), a the acceleration and t the time.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 7, 2013
Today on TSR

### Tuition fees under review

Would you pay less for a humanities degree?

### Buying condoms for the first time - help!

Discussions on TSR

• Latest
Poll
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE