Question

A 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, held in place by static friction that can be at most 3.6 N. The car starts from rest and gradually accelerates on a straight road. How fast is the car moving when the pebble flies out of the tire tread?

Answer 1

When a 1.2 g pebble is stuck in a tread of a 0.76 m diameter automobile tire, the car is moving approximately 33.74 m/s when the pebble flies out of the tire tread.

Step-by-step explanation:

To find the speed of the car when the pebble flies out of the tire tread, we need to consider the forces acting on the pebble and the car.

The maximum static friction that can hold the pebble in place is given as 3.6 N.

When the car accelerates, a centripetal force is exerted on the pebble due to its circular motion in the tire tread.

The centripetal force can be calculated using the formula F = mv²/r

Where

m is the mass of the pebble

v is the velocity of the cae

r is the radius of the tire.

Equating the centripetal force to the maximum static friction force, we can solve for the velocity of the car.

Let's calculate the centripetal force first. The mass of the pebble is given as 1.2 g = 0.0012 kg.

The radius of the tire is half of its diameter, so it is 0.76 m/2 = 0.38 m.

Plugging these values into the formula, we get F = (0.0012 kg)(v²/0.38 m).

Equating the centripetal force to the maximum static friction force, we have (0.0012 kg)(v²/0.38 m) = 3.6 N.

Solving for v, we get v² = (3.6 N)(0.38 m)/(0.0012 kg), which simplifies to v² = 1140 m²/s².

Taking the square root of both sides, we find v = sqrt(1140 m²/s²) ≈ 33.74 m/s.

Answer 2

`v=33.764m/s`

Step-by-step explanation:

Given data

Mass m=1.2 g=0.0012 kg

diameter d=0.76 m

Friction Force F=3.6 N

To find

Velocity v

Solution

From the Centripetal force we know that

`F_(c)=(mv^(2) )/(r)`

Where m is mass

v is velocity

r is radius

Substitute the given values to find velocity v

So

`F_(c)=(mv^(2) )/(r)\\v^(2)=(F_(c)(r))/(m)\\ v=\sqrt{(F_(c)(r))/(m)}\\ v=\sqrt{(F_(c)(diameter/2))/(m)}\\v=\sqrt{((3.6N)(0.76/2)m)/((0.0012kg))}\\v=33.764m/s`