A 1-kg thin hoop with a 50-cm radius rolls down a 47° slope without slipping. If the hoop starts from rest at the top of the slope, what is its translational velocity after it rolls 16 m along the slope?
I got the height by 16sin47=11.7 m
I tried putting the k.e and p.e equations equal......lost!
Step 1: We know that energy is conserved... so we know that KE(initial)+ PE(initial)= KE(final)+ PE (final).
Step 2: Since the hoop is initially at rest at the top of the hill, KE (initial) is 0. And after the hoop travels down the hill, all the PE is converted to KE, so PE(final) is equal to 0. So now we have that PE(initial)=KE(final)
Step 3: Plug in the formulas for PE and KE. PE is equal to mass*gravity*change in height. KE in this problem is KE(rotational)+KE(transitional). KE(rotational)= (1/2)*moment of inertia*angular velocity^2. KE(transitional)= (1/2)*mass*velocity^2.
So know we have: m*g*change in height= (1/2)*I*w^2+ (1/2)*m*v^2
Step 4: angular velocity (w) is equal to tangential velocity divided by radius (v/r). So substituting in this value for w, we now have: m*g*change in height= (1/2)*I*(v/r)^2+ (1/2)*m*v^2
Step 5: The moment of inertia for a hoop is mass*radius^2. So substituting this in for I, we now have: m*g*change in height=(1/2)*m*(r^2)*(v/r)^2+ (1/2)*m*v^2
Step 6: Simplify the equation! Notice that all the terms are multiplied by a factor of m, so we can pull it out. Also notice that in the KE(rotational) term r^2 cancels out. So now we have: g*change in height= (1/2)*v^2+(1/2)*v^2
Step 7: Solve for change in height! We know that the slope is 16 meters. And the angle of the slope is 47. sin(theta)=(opposite/hypotenuse)... so the change in height is 16*sin(47). Plugging that in for change in height, we now have: g*16*sin(47)= (1/2)*v^2+(1/2)*v^2
Step 8: Plug in 9.8 for gravity and solve for v. We get that v= 10.7 m/s!!!!
Good Luck!!! :)
To calculate the hoop's translational velocity after it rolls 16 m along the slope, you need to consider the potential energy and the kinetic energy of the hoop.
First, let's calculate the potential energy (PE) at the top of the slope. The potential energy is given by the equation: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the mass (m) is 1 kg, the acceleration due to gravity (g) is 9.8 m/s^2, and the height (h) is given as 11.7 m.
PE = (1 kg) × (9.8 m/s^2) × (11.7 m)
PE = 114.54 Joules
Next, let's consider the kinetic energy (KE) of the hoop after rolling 16 m along the slope. The kinetic energy is given by the equation: KE = 0.5mv^2, where m is the mass and v is the velocity.
Since the hoop is rolling without slipping, the relationship between the translational velocity (Vt) and the angular velocity (ω) is given by Vt = R * ω, where R is the radius of the hoop.
In this case, the radius (R) is given as 50 cm, which is equal to 0.5 m. To find the angular velocity (ω), we need to use the relationship between the linear velocity (Vl) and the angular velocity, which is Vl = R * ω. Since the hoop is rolling without slipping, the linear velocity at the top of the slope is zero.
Vl = 0.5 m * ω
0 m/s = 0.5 m * ω
ω = 0 rad/s
So, the initial angular velocity is zero. Now, let's calculate the final angular velocity (ω) using the equation: ω = v / R, where v is the final velocity.
ω = v / 0.5 m
ω = 2v rad/s
The kinetic energy (KE) in terms of the angular velocity is given by the equation: KE = 0.5Iω^2, where I is the moment of inertia of the hoop. For a thin hoop, the moment of inertia is I = mR^2.
KE = 0.5 * (1 kg) * (0.5 m)^2 * (2v rad/s)^2
KE = 0.25 kg m^2/s^2 * (2v rad/s)^2
KE = 0.25 kg m^2/s^2 * 4v^2 rad^2/s^2
KE = 1 kg m^2/s^2 * v^2 rad^2/s^2
KE = v^2 Joules
Now, we can equate the potential energy (PE) to the kinetic energy (KE) to solve for the final velocity (v):
114.54 J = v^2
v = √(114.54 J)
v = 10.71 m/s
Therefore, the hoop's translational velocity after it rolls 16 m along the slope is approximately 10.71 m/s.
To calculate the translational velocity of the hoop, we need to consider the conservation of energy. We'll start by finding the potential energy at the starting point and the kinetic energy as the hoop rolls down.
1. Find the potential energy at the starting point:
The potential energy (PE) is given by the formula PE = m * g * h, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the height is the vertical distance the hoop has to travel, which is 11.7 m.
Since the hoop has a mass of 1 kg and the acceleration due to gravity is approximately 9.8 m/s², we can calculate the potential energy:
PE = 1 kg * 9.8 m/s² * 11.7 m = 113.46 J
2. Find the kinetic energy at the bottom of the slope:
As the hoop rolls down the slope without slipping, it gains both rotational kinetic energy and translational kinetic energy. Since the hoop starts from rest at the top of the slope, the initial kinetic energy is zero.
To calculate the final kinetic energy, we can consider the hoop as a point particle rotating around its center of mass. The equation for the kinetic energy (KE) of an object rotating about its center of mass is given by KE = (1/2) * I * ω², where I is the moment of inertia and ω is the angular velocity.
For a thin hoop, the moment of inertia is given by I = m * r², where m is the mass and r is the radius. In this case, m = 1 kg and r = 0.5 m (50 cm).
I = 1 kg * (0.5 m)² = 0.25 kg·m²
The angular velocity ω can be calculated using the relation between linear velocity and angular velocity:
v = ω * r, where v is linear velocity and r is the radius.
We can rearrange the equation to solve for ω:
ω = v / r
Now, we have the linear velocity v and the radius r, so we can calculate the angular velocity ω.
3. Calculate the final angular velocity:
To find the final angular velocity, we need to relate the distance traveled by the hoop to its angular displacement. As the hoop rolls down the ramp without slipping, for every distance of 2πr traveled by the hoop, it completes one full revolution.
The linear distance traveled along the slope is 16 m, which corresponds to an angular displacement of (16 m / (2π * 0.5 m)) * 2π radians.
Now we can calculate the final angular velocity ω:
ω = (angular displacement) / (time)
ω = [16 m / (2π * 0.5 m)] * 2π radians / (time taken to cover the distance of 16 m)
4. Calculate the translational velocity:
To find the translational velocity, we can use the relation between angular velocity and translational velocity:
v = ω * r
Now, we have the angular velocity ω and the radius r, so we can calculate the translational velocity.
By following these steps, you should be able to calculate the translational velocity of the hoop after it rolls 16 m along the slope.